LIGHT:REFLECTION AND REFRACTION-III : LONG ANSWERS

GIVE ANSWER IN DETAIL

*INDEX-TOPIC SEARCH

1. EXPLAIN WITH DIAGRAM : REFLECTION FROM PLANE MIRROR.
   
   
   

   ANS:
   
   Consider an object AO of height h placed in front of a plane mirror MM' at a distance u.
   
   Each small portion of this extended object AO acts like a point source.
   
   Consider two incident rays AN and AQ from the endpoint A of the object AO.
   
   According to the law of reflection, NA and QR are corresponding reflected rays.
   
   Since ray AN is normal to the mirror, the ray NA is reflected back on the same path.
   
   Similarly, OQ is normal to the mirror. Therefore QO is reflected back on the same path.
   
   Rays NA and QR are reflected rays which meet after extending backwards (behind the mirror) at A'. Thus A' is the virtual image of A. Similarly all the points of object AO form their virtual image behind the mirror. Thus the image A'I of all the points of AO is formed at a distance v behind the mirror. This image is virtual, erect and of the same size (height) as the object. The distances u and v are equal,i.e. the image is at the same distance behind the mirror as the object is in front of the mirror.
   The image is laterally inverted,i.e.,if a person raises the left hand the image appears to raise the right hand.
   



2. EXPLAIN WITH DIAGRAM : CARTESIAN SIGN CONVENTION FOR SPHERICAL MIRROR.
   

   ANS:
   

   
   
   • The pole P of the mirror is taken as the origin.
     
     
   • The principal axis of the mirror is taken along X-axis.
     
     
   • The object is considered as placed on the left of the mirror, i.e. the light is incident on the mirror from the left hand side of the mirror.
     
     
   • All the distances along the principal axis are measured from the pole (P).
     
     
     
   • The distances measured in the direction of the incident ray are taken as positive.
     
     
   • The distances measured in the direction opposite to the direction of incident ray are taken as negative.
     
     
   • The heights measured upwards and perpendicular to the principal axis of the mirror are taken as positive.
     
     
   • The heights measured downwards and perpendicular to the principal axis of the mirror are taken as negative.
     

   
   



3. DERIVE MIRROR FORMULA.
   

   ANS:
   
   The relation between object distance(u), image distance(v) and the focal length (f) of the mirror is known as MIRROR FORMULA.
   
   
   Consider a concave mirror MM' (of small aperture).
   
   An object AB of height h is placed on the left of the mirror beyond its center of curvature so that the image formed is real, inverted and diminished in size(height = h').
   
   Here :
   
   Object distance = PB = -u
   
   Image distance = PB'= -v
   
   Focal length = PF = -f
   
   Radius of curvature = PC = -R
   
   As shown in Figure, Δ ABP ∼ Δ A'B'P.
   
   ∴ A'B'/AB = PB'/PB
   
   ∴ A'B'/AB = (-v)/(-u)
   
   ∴ A'B'/AB = v/u.............(I)
   
   Similarly, as shown in the figure, Δ ABC ∼ Δ A'B'C.
   
   ∴ A'B'/AB = CB'/CB
   
   ∴ A'B'/AB = (PC-PB')/(PB-PC)........[as CB'=(PC-PB') and CB=(PB-PC)]
   
   ∴ A'B'/AB = [-R - (-v)]/[-u - (-R)]
   
   ∴ A'B'/AB = (-R + v)/(-u + R)........(II)
   
   By (I) & (II),
   
   v/u = (-R + v)/(-u + R)
   
   ∴ -uv + vR = -uR + uv
   
   ∴ vR + uR = 2uv
   
   ∴ (vR/uvR) + (uR/uvR) =2uv/uvR.......[Dividing each side by uvR]
   
   ∴ (1/u) + (1/v) = (2/R)..............(III)
   
   Now, for an object placed at infinite distance (u = ∞), the image is formed at the focus F (v = f).Thus, eq.(III) reduces to
   
   (1/∞) + (1/f) = (2/R)
   
   ∴ (2/R) = (1/f)....................(IV)
   
   comparing eq.(III) and (IV),
   
   (1/u) + (1/v) = (1/f)
   



4. EXPLAIN : MAGNIFICATION AND DERIVE ITS FORMULA.
   

   ANS:
   The ratio of the height of the image to the height of the object is called MAGNIFICATION.
   
   Magnification, m = image height/object height = h'/h
   
   From the figure(see the figure in derivation of MIRROR FORMULA), it is clear that
   
   Δ ABP ∼ Δ A'B'P
   
   ∴ A'B'/AB = PB'/PB

   A'B' = image height = h'
   AB = object height = h
   PB = object distance = u
   PB' = image distance = v

   
   
   ∴ h'/h = u/v
   
   ∴ m = u/v
   
   According to the sign convention, m = (-h')/(h).
   
   ∴ m = -(u/v).
   
   The object height(h) is always positive. The image height(h') is positive when the image is virtual and erect whereas h' is negative when the image is real and inverted.
   
   For a plain mirror,
   
   image height(h') = object height(h)
   
   ∴ m = (+1).
   
   If m > 1, the image is enlarged(actual magnification)
   
   If m = 1, the object and image are of same size
   
   If m < 1, the image is smaller than the object.
   
   Magnification is negative for real images and positive for virtual images. Thus, m is always positive for convex mirrors and it is negative for the real images formed by a concave mirror.
   



5. EXPLAIN : POWER OF LENS.
   

   ANS:
   

   
   
   • The power of lens is the reciprocal of its focal length (f).
     
     ∴ Power p = 1/f
     
     
   • The S. I. unit of power of a lens is dioptre (symbol D)
     
     Thus, 1 D = 1 m^-1
     
     
   • The power of a convex lens is positive(+) and that of a concave lens is negative(-).
     
     
   • If the focal length of a convex lens is 0.5 m, then its power is
     
     1/0.5 or 2.0 D.
     
     
   • If the focal length of a concave lens is (-0.2 m), then its power is
     
     -(1/0.2) or -5.0 D.
     
     
   • An ophthalmologist (or opticion), after examining a patient, prescribes corrective lenses for eyeglasses or spectacles in terms of dioptre (D).
     
     
   • Dioptremeter is used to measure the power of a lens.
     

   
   

   
   
6. State Laws of refraction and derive generalised form of Snell's law.
   

   ANS:
   The laws of refraction are:
   
   

   • The ratio of the sine of the angle of incidence(θ₁) to the sine of the angle of refraction(θ₂) is constant.
     
   • The incident ray, the normal to the surface separating two mediums and the refracted ray, all lie in the same plane.

   
   
   
   As shown in the figure,surface AB separates medium 1 and medium 2. PQ is the ray incident on surface AB at point Q, QR is the refracted ray and NQN' is the normal to the surface AB at Q.
   
   ∠ PQN = incident angle = θ₁
   
   ∠ RQN'= refraction angle = θ₂
   
   According to the law of refraction,
   
   sin θ₁ / sin θ₂ = constant
   
   The constant in the above equation is called the refractive index of medium 2 with respect to medium 1 and is denoted by η₂₁
   
   ∴ η₂₁ = sin θ₁ / sin θ₂
   
   The above equation is called Snell's law.The value of η₂₁ depends upon the nature of the two mediums.
   
   Now,
   
   η₂₁ = vel. of light in medium 1/vel. of light in medium 2
   
   ∴ η₂₁ = v₁/v₂
   

   [here,v₁ = velocity of light in medium 1
   v₂ = velocity of light in medium 2]

   
   
   Also, by the definition of absolute refractive index,
   
   Absolute refractive index of medium 1 =
   

   velocity of light in vaccum
   velocity of light in medium 1
   

   
   
   ∴ η₁ = c/v₁
   
   Similarly,
   
   the absolute refractive index of medium 2, η₂ = c/v₂.
   
   ∴ (η₂/η₁) = (v₁/v₂)
   
   ∴ (η₂/η₁) = η₂₁
   
   ∴ (η₂/η₁) = sin θ₁ / sin θ₂
   
   ∴η₁ sin θ₁ = η₂ sin θ₂
   
   This is called generalised form of SNELL'S LAW.
   



7. Derive LENS FORMULA.
   

   ANS:
   
   The relation between object distance(u), image distance(v) and the focal length(f) of a lens is known as LENS FORMULA.
   
   
   Consider a (thin spherical) convex lens MN (of small aperture).
   
   An object AB of height h is placed in front of the lens MN beyond 2F on its left side.
   
   The image A'B' formed by the lens is real, inverted and diminished in size (h').
   
   Here,
   

   Object Distance = OB = -u
   
   Image Distance = OB' = +v
   
   Focal Length = OF = +f

   
   
   As shown in figure, Δ ABO ∼ Δ A'B'O
   
   ∴ (AB/A'B') = (OB/OB')
   
   ∴ (AB/A'B') = (-u/v) _______________(I)
   
   Similarly, Δ OCF ∼ Δ A'B'F
   
   ∴ (OC/A'B') = (OF/FB')
   
   ∴ (AB/A'B') = (OF/FB')........(as AB = OC)
   
   ∴ (AB/A'B') = (OF)/(OB' - OF)
   
   ∴ (AB/A'B') = (f)/(v - f) __________(II)
   
   From Eq. (I) and (II), we have
   
     (-u/v) = (f)/(v - f)
   
   ∴ -u(v - f) = vf
   
   ∴ -uv + uf = vf
   
   ∴ (-uv/uvf) + (uf/uvf) = (vf/uvf)...(dividing by uvf)
   
   ∴ (-1/f) + (1/v) = (1/u)
   
   ∴ (1/f) = (1/v) - (1/u)
   
   This is known as LENS FORMULA.
   



8. EXPLAIN : TWINKLING OF STARS
   

   ANS:
   

   
   
   
   • The density of the atmosphere is not uniform everywhere as the atmosphere consists of layers of different densities.
     
     
   • The layer at the lower altitude (near earth's surface) is colder and denser than the layer at higher altitude.
     
     
   • Due to this , the refractive index of atmosphere decreases continuously as we move up from the surface.
     
     
   • The star light keeps on bending towards the normal at every point of the atmosphere till it reaches our eyes.
     
     
   • Thus, as shown in the figure(refer to the figure given in the answer for TOTAL INTERNAL REFLECTION) the position of the star appears different from its actual position.
     
     
   • Moreover, the atmosphere is not stationary, i.e. the physical condition of the refractive medium keeps on changing every moment.
     
     
   • This results in continuous shifting of the apparent position of the star as well as variation in the intensity of its light.
     
     
   • This leads to TWINKLING OF STARS.
     
     
   • Unlike stars the planets do not twinkle because they are far closer than stars. Therefore, we can consider stars as point-sized source of light and planets as a collection of large number of point-sized sources (extended objects) of light such that they nullify twinkling effect.
     

   
   
9. EXPLAIN : TOTAL INTERNAL REFLECTION (MIRAGE).
   

   ANS:
   

   
   
   
   • When a ray of light travels from optically denser medium to optically rarer medium it bends away from the normal at the surface separating two mediums due to refraction.
     
     
   • As the angle of incidence(θ_i) increases, the angle of refraction(θ_r) also increases and the ray of light moves further away from the normal.
     
     
   • For some value of angle of incidence, the angle of refraction becomes 90°.The angle of incidence for which the angle of refraction is 90°, is called CRITICAL ANGLE.
     
     
   • If the angle of incidence is greater than the critical angle, the angle of refraction becomes greater than 90° and the refracted ray is confined to the same medium only.This is calledTOTAL INTERNAL REFLECTION.
     
     
     
   • MIRAGE:
     
     
     
     
     • During summer, the air near the surface of the earth is hotter than the air above it. Therefore, the air near the surface of the earth is rarer than the air above it. Thus, as we move up from the surface of the earth, the refractive index of air increases.
       
       
     • As shown in figure when the light rays coming from the top point (P) of an object (like tree) pass thruogh different layers of air up to the ground, they move more and more away from the normal.
       
       
     • Thus the angle of refraction will go on increasing and the ray will undergo total internal reflection when it reaches the ground. This totally reflected ray enters our eye and the image of P is seen as P' as if reflection has occurred due to water.
       This is referred to as mirage
       
     
     

   
   
   
10. EXPLAIN : REFRACTION THROUGH A RECTANGULAR SLAB.
    

    ANS:
    
    
    As shown in figure, let []PQRS be a rectangular slab.
    
    A ray of light AB is incident at point B on surface PQ at an angle θ₁.
    
    At point B this ray is refracted making an angle θ₂ and travels as ray BC in glass.
    
    Ray BC is incident on surface RS at point C making an angle θ₃.
    
    From point C, the ray CD emerges from rectangular slab making an angle θ₄.
    
    If η₁,η₂ and η₃ are taken refractive indices of air, glass and air, we have
    
    η₁ = η₃ = 1
    
    According to the laws of refraction, at surface PQ,
    
      η₂ sin θ₂ = η₁ sin θ₁
    
    ∴ η₂ sin θ₂ = sin θ₁......(as η₁ = 1).......(I)
    
    At the surface RS parallel to PQ,
    
      η₂ sin θ₃ = η₃ sin θ₄
    
    ∴ η₂ sin θ₂ = η₃ sin θ₄......(as θ₂ = θ₃)
    
    ∴
    η₂ sin θ₂ = sin θ₄......(as η₃ = 1).......(II)
    
    By (I) and (II),
    
      sin θ₁ = sin θ₄
    
    ∴ θ₁ = θ₄
    
    Thus, the direction of the emergent ray CD is parallel to incident ray AB(EF) but because of the refraction, emergent ray is displaced by perpendicular distance CE. This kind of displacement (CE) is known as LATERAL SHIFT. The lateral shift is proportional to the perpendicular distance between two refracting surfaces.
    

    
    
11. DISTINGUISH BETWEEN CONVERGING AND DIVERGING LENS.
    

    ANS:
    

    
    

LIGHT:REFLECTION AND REFRACTION-II:BRIEF ANSWERS

ANSWER IN BRIEF

*INDEX-TOPIC SEARCH

1. EXPLAIN THE NATURE OF LIGHT IN BRIEF
   

   ANS:
   

   • Light is an electromagnetic radiation in the form of waves and it does not require material medium for its propagation.
   • Light travels at the speed of 3 x 10⁸ m/s in vacuum.
   • When it passes through a transparent medium its speed reduces significantly.
   • The wavelength of visible light is 4 x 10^(-7) to 8 x 10^(-7) m.
   • Light incident on a surface separating two media is partly reflected, partly refracted(transmitted), and partly absorbed.
   • Highly polished surface reflects a large amount of incident light whereas a transparent medium refracts(transmits) most of the light incident on it.
   • A lens or mirror focuses light falling on it because of reflection or refraction.
   • Due to reflection by a mirror or refraction by a lens, the image of an object is formed or it appears to have formed.
   
   



2. EXPLAIN:REGULAR AND IRREGULAR REFLECTION
   

   ANS:When light is incident on an opaque surface,reflection occurs.There are two types of reflection:(i)Regular reflection and (ii)Irregular reflection.
   
   
   
   Regular Reflection:When a parallel beam of light falls on a smooth or highly polished (opaque) surface, the reflected beam is also parallel and directed in a fixed direction.This type of reflection is called regular reflection.For example,reflection by a mirror. Regular reflection does not enable us to see the object.
   
   Irregular Reflection:When a parallel beam of light falls on a rough surface, the reflected beam is not parallel and it spreads over a wide area. This is called irregular reflection.Irregular reflection enables us to see the object.
   



3. STATE LAWS OF REFLECTION
   

   ANS:The laws of reflection are:
   
   

   • The angle of incidence(θ_i) is equal to the angle of reflection(θ_r).
   
   
   • The incident ray, the normal to the mirror at the point of incidence and the reflected ray, all lie in the same plane.
   
   



4. EXPLAIN : Real image and virtual image.
   

   ANS:
   
   

   
   
5. MENTION CHARACTERISTICS OF IMAGE FORMED BY A PLANE MIRROR.
   

   ANS:

   
   
   • The image is virtual, erect and of the same size as the object.
   • The image is at the same distance behind the mirror as the object is in front of the mirror.
   • The image is laterally inverted,i.e., when a person raises the left hand the image appears to raise the right hand.
     

   
   

   
   
6. Draw the ray diagram to show the position, nature and the size of the image when object is placed at infinity from a concave mirror.
   

   ANS:
   
   
   
   The image is formed at focus(F); it is real, inverted and highly diminished.



7. Draw the ray diagram to show the position, nature and the size of the image when object is placed beyond C (Center of curvature) of a concave mirror.
   

   ANS:
   
   
   
   The image is formed between focus(F) and center of curvature(C); it is real, inverted and diminished.



8. Draw the ray diagram to show the position, nature and the size of the image when object is placed at C (Center of Curvature)of a concave mirror.
   

   ANS:
   
   
   
   The image is formed at C; it is real, inverted and of the same size as the object.



9. Draw the ray diagram to show the position, nature and the size of the image when object is placed between C and F of a concave mirror.
   

   ANS:
   
   
   
   The image is formed beyond C; it is real, inverted and enlarged.



10. Draw the ray diagram to show the position, nature and the size of the image when object is placed at focus(F) of a concave mirror.
    

    ANS:
    
    
    
    The image is formed at infinite distance; it is real, inverted and highly enlarged.



11. Draw the ray diagram to show the position, nature and the size of the image when object is placed between pole(P) and F of a concave mirror.
    

    ANS:
    
    
    
    The image is formed behind the mirror; it is virtual and erect and enlarged.



12. Draw the ray diagram to show the position, nature and the size of the image when object is placed at infinity from a convex mirror.
    

    ANS:
    
    
    
    The image is formed at the focus F behind the mirror; it is virtual and erect and highly diminished.



13. Draw the ray diagram to show the position, nature and the size of the image when object is placed between pole(P) of a convex mirror and infinity.
    

    ANS:
    
    
    
    The image is formed between P and F behind the mirror; it is virtual and erect and diminished.



14. Draw the ray diagram to show the position, nature and the size of the image when object is placed at infinity from a convex lens.
    

    ANS:
    
    
    The image is formed on the opposite side of lens at focus (F₂); it is real and inverted and highly diminished(point-sized).



15. Draw the ray diagram to show the position, nature and the size of the image when object is placed beyond 2F₁ of a convex lens.
    

    ANS:
    
    
    
    The image is formed between F₂ and 2F₂of the lens; it is real, inverted and diminished.



16. Draw the ray diagram to show the position, nature and the size of the image when object is placed at 2F₁ of a convex lens.
    

    ANS:
    
    
    
    The image is formed at 2F₂; it is real inverted and of same size as the object.



17. Draw the ray diagram to show the position, nature and the size of the image when object is placed between F₁ and 2F₁ of a convex lens.
    

    ANS:
    
    
    
    The image is formed beyond 2F₂; it is real, inverted and enlarged.



18. Draw the ray diagram to show the position, nature and the size of the image when object is placed at F₁ of a convex lens.
    

    ANS:
    
    
    
    The image is formed at infinity; it is real, inverted and highly enlarged.



19. Draw the ray diagram to show the position, nature and the size of the image when object is placed between F₁ and optical center O of a convex lens.
    

    ANS:
    
    
    
    The image is formed on the same side of the lens as the object; it is virtual,erect and enlarged.



20. Draw the ray diagram to show the position, nature and the size of the image when object is placed at infinity from a concave lens.
    

    ANS:
    
    
    
    The image is formed at the focus F₁ on the same side of the object; it is virtual, erect and highly diminished.



21. Draw the ray diagram to show the position, nature and the size of the image when object is placed between infinity and the optical center O of a concave lens.
    

    ANS:
    
    
    
    The image is formed between focus F₁ and the optical center O on the same side of the object; it is virtual, erect and diminished.



22. STATE LAWS OF REFRACTION.
    

    ANS:The laws of refraction are:
    
    

    • The ratio of the sine of the angle of incidence(θ₁) to the sine of the angle of refraction(θ₂) is constant.
    
    
    • The incident ray, the normal to the surface separating two mediums and the refracted ray, all lie in the same plane.
    
    



23. DISTINGUISH BETWEEN REFLECTION AND REFRACTION.
    

    ANS:
    
    



24. STATE SNELL'S LAW AND WRITE ITS FORMULA.
    

    ANS:The ratio of the sine of the angle of incidence and the sine of the angle of refraction is always constant for given pair of mediums.
    
    sinΘ₁/sinΘ₂ = constant
    
    The constant is known as the refractive index of medium 2 with respect to medium 1, and is denoted by η₂₁

    
    
25. THE REFRACTIVE INDEX OF TWO MEDIUMS A AND B ARE 1.33 AND 1.5, RESPECTIVELY.
    1. IF A RAY TRAVELS FROM A TO B, WHAT WILL HAPPEN TO ITS SPEED?
    2. WILL TOTAL INTERNAL REFLECTION OCCUR IF A RAY TRAVELS FROM A TO B?
       
    
    

    ANS:

    
    
    1. The speed of the light ray will decrease as it enters a denser medium from a rarer medium.
    2. No, total internal reflection will not occur because the ray will move towards the normal in this case.
       

    
    

LIGHT:REFLECTION AND REFRACTION-I:MCQs AND SHORT ANSWERS

MULTIPLE-CHOICE QUESTIONS

*INDEX-TOPIC SEARCH

SELECT THE CORRECT ALTERNATIVE:

1. The speed of light is_______ in vacuum.
   
   
   
   1. 3 x 10⁵ m/s
      
   2. 3 x 10⁸ m/s
      
   3. 3 x 10⁸ km/s
      
   4. 3 x 10⁶ m/s
      
   
   
2. The wavelength of the visible light is ________.
   
   
   
   1. 4 x 10^(-7) to 8 x 10^(-7) m
      
   2. 4 x 10⁷ to 8 x 10⁷ m
      
   3. 4 x 10^(-7) to 8 x 10^(-7) Å
      
   4. 4 x 10⁷ to 8 x 10⁷ Å
      
   
   
3. We can see objects because of_______
   
   
   
   1. reflection
      
   2. refraction
      
   3. transmission
      
   4. diffraction
      
   
   
4. The image formed by a convex mirror is always________
   
   
   
   1. real
      
   2. enlarged
      
   3. virtual and enlarged
      
   4. diminished
      
   
   
5. As you move an object away from a convex mirror, its image becomes_____ and moves towards______
   
   
   
   1. smaller, infinity
      
   2. smaller, focus
      
   3. enlarged, infinity
      
   4. enlarged, focus
      
   
   
6. For a spherical mirror, ______ is true.
   
   
   
   1. f = 2R
      
   2. R = 2f
      
   3. fR = 2
      
   4. fR = 1/2
      
   
   
7. The mirror formula is ________.
   
   
   
   1. 1/u - 1/v = 1/f
      
   2. 1/f + 1/u = 1/v
      
   3. f = uv/(u+v)
      
   4. f = (u+v)/uv
      
   
   
8. For a plane mirror, magnification (m)=________
   
   
   
   1. 0
      
   2. 1
      
   3. ± 1
      
   4. ≤ 0
      
   
   
9. Magnification for convex mirror is ________.
   
   
   
   1. always positive
      
   2. always negative
      
   3. sometimes positive
      
   4. 1
      
   
   
10. The image formed by a concave lens is ________.
    
    
    
    1. always real and enlarged
       
    2. always real and diminished
       
    3. always virtual and enlarged
       
    4. always virtual and diminished
       
    
    
11. The lens formula is ________.
    
    
    
    1. 1/f = 1/u + 1/v
       
    2. 1/f = 1/u - 1/v
       
    3. 1/f = 1/v - 1/u
       
    4. 1/f + 1/v = u
       
    
    
12. 1 diopter = ________.
    
    
    
    1. 1 m
       
    2. 1 m^(-1)
       
    3. 1 cm
       
    4. 1 cm^(-1)
       
    
    
13. Which of the following is a true statement?
    
    
    
    1. The power of a lens is always positive.
       
    2. The power of a lens is always negative.
       
    3. The power of a convex lens is positive.
       
    4. The power of a concave lens is positive.
       
    
    
14. Image formed by a concave mirror is erect and enlarged.What is the position of the object?
    
    
    
    1. Between focus F and the center of curvature
       
    2. At the center of curvature
       
    3. Beyond the center of curvature
       
    4. Between pole and the focus
       
    
    
15. If the focal length of a spherical mirror is 40 cm, then its radius of curvature is _______ cm.
    
    
    
    1. 80
       
    2. 20
       
    3. 10
       
    4. 5
       
    
    
16. The velocity of light in vacuum is _______ ms^(-1)
    
    
    
    1. 3 x 10⁶
       
    2. 3 x 10⁸
       
    3. 3 x 10 ¹²
       
    4. 3 x 10¹⁵
       
    
    
17. If the angle of incidence, θ_i = 0°, the angle of reflection, θ_r = ________.
    
    
    
    1. 0°
       
    2. 90°
       
    3. 180°
       
    4. 45°
       
    
    
18. No matter how far is the object from the mirror, the image of the object appears erect. The mirror is ________.
    
    
    
    1. concave
       
    2. convex
       
    3. either concave or convex
       
    4. none of these
       
    
    
19. A boy is standing at a distance of 2 m in front of a plane mirror.The distance between the boy and his image is ________ m.
    
    
    
    1. 4
       
    2. 3
       
    3. 2
       
    4. 1
       
    
    
20. The image formed by a concave mirror is real, inverted and of the same size as that of the object.The position of the object should be ________.
    
    
    
    1. beyond the center of curvature of mirror
       
    2. between the center of curvature and the focus
       
    3. at the center of curvature of the mirror
       
    4. at the focus
       
    
    
21. Which of the following has the highest refractive index?
    
    
    
    1. Glass
       
    2. Water
       
    3. Pearl
       
    4. Diamond
       
    
    
22. Absolute refractive index of any medium is always _______.
    
    
    
    1. 1
       
    2. > 1
       
    3. < 1
       
    4. 0
       
    
    
23. The image formed by a plane mirror is ______.
    
    
    
    1. real
       
    2. diminished
       
    3. enlarged
       
    4. laterally inverted
       
    
    
24. The incident ray passing through the focus(F) of a mirror ______ after reflection.
    
    
    
    1. passes through C
       
    2. passes through F
       
    3. becomes parallel to the principal axis
       
    4. passes through the pole
       
    
    
25. The incident ray passing through the center of curvature(C) of a mirror ______ after reflection.
    
    
    
    1. passes through C
       
    2. passes through F
       
    3. passes through the pole
       
    4. becomes parallel to the principal axis
       
    
    
26. The incident ray parallel to the principal axis of a mirror ______ after reflection.
    
    
    
    1. passes through C
       
    2. passes through F
       
    3. passes through the pole
       
    4. reverts back in the opposite direction
       
    
    
27. According to the sign convention, the distance of object...
    
    
    
    1. is always positive
       
    2. is always negative
       
    3. may be positive or negative
       
    4. is equal to object height
       
    
    
28. According to the sign convention, the distance of image...
    
    
    
    1. is always positive
       
    2. is always negative
       
    3. may be positive or negative
       
    4. is equal to image height
       
    
    
29. The refractive index of a denser medium with respect to a rarer medium is...
    
    
    
    1. 1
       
    2. greater than 1
       
    3. smaller than 1
       
    4. negative
       
    
    
30. The refractive index of a rarer medium with respect to a denser medium is...
    
    
    
    1. 1
       
    2. greater than 1
       
    3. smaller than 1
       
    4. negative
       
    
    
31. Total internal reflection will occur if the angle of reflection is...
    
    
    
    1. 45°
       
    2. 60°
       
    3. 90°
       
    4. 99°
       
    
    
32. Magnification for ______ image is always ______.
    
    
    
    1. real, positive
       
    2. real, negative
       
    3. virtual, negative
       
    4. any, negative
       
    
    
33. If magnification is +1.5, the image is______.
    
    
    
    1. erect
    2. diminished
    3. real
    4. inverted
       
    
    

ANSWERS TO MCQs:

(1) B (2) A (3) A (4) D (5) B (6) B (7) C (8) B (9) A (10) D (11) C (12) B (13) C (14) D (15) A (16) B (17) A (18) B (19) A (20) C (21) D (22) B (23) D (24) C (25) A (26) B (27) B (28) C (29) B (30) C (31) D (32) B (33) A

SHORT ANSWERS

1. What is light?
   
   

   ANS:Light is an electromagnetic radiation which produces sensation in our eyes.

   
   
2. What is the range of wavelength of visible light in Å units ?
   

   ANS:4000 to 7000 Å
   

   
   
3. Which waves do not require material medium for propagation?
   
   

   ANS: Electromagnetic waves do not require material medium for propagation.

   
   
4. What happens to the speed of light when it passes through a transparent medium?
   
   

   ANS: The speed of light is reduced significantly when it passes through a transparent medium.

   
   
5. What happens to light when it is incident on the surface separating two mediums ?
   

   ANS:A part of the incident light is reflected, a part is transmitted and rest is absorbed.
   

   
   
6. Define: Ray of light.
   

   ANS: A straight line path joining one point to another in the direction of propagation of light is known as a ray.

   
   
7. What is a beam of light?
   

   ANS: A bundle of light rays is called a beam of light.

   
   
8. Which phenomena enable the focusing of light by lens and mirrors?
   

   ANS:Refraction and reflection are the phenomena which enable the focusing of light by lens and mirror.

   
   
9. What is an image?
   

   ANS:Whan a number of rays starting from a point meet at another point after reflection or refraction, the second point is called the image of the first point.

   
   
10. Define: Real image.
    

    ANS:When the rays emerging from one point meet really at another point after reflection or refraction, the image formed is called real image.

    
    
11. What is virtual image?
    

    ANS:When the rays emerging from one point appear to meet, when extended backwards, after reflection or refraction, the image formed is virtual image.

    
    
12. Mention two kinds of reflection.
    

    ANS:Regular reflection and Irregular(diffused) reflection are the two kinds of reflection.

    
    
13. Which type of surfaces produce regular reflection?
    

    ANS:Smooth and highly polished surfaces produce regular reflection.

    
    
14. Which type of surface produces irregular reflection?
    

    ANS:Rough surface produces irregular reflection.

    
    
15. What is 'incident angle'?
    

    ANS:The angle that the incident ray makes with the normal is known as incident angle(or Angle of incidence).

    
    
16. What is 'reflection angle'?
    

    ANS:The angle that the reflected ray makes with the normal is known as reflection angle (or Angle of reflection).

    
    
17. How are curved mirrors formed?
    

    ANS:Curved mirrors are formed by cutting circular cross-section of spherical shell.

    
    
18. Define: Radius of curvature of a mirror.
    

    ANS:The radius of the spherical shell from which the mirror is made is called the radius of curvature of a curved mirror.

    
    
19. Define: Centre of curvature of a mirror.
    

    ANS:The center of the spherical shell from which the mirror is made is called the centre of curvature of the mirror.

    
    
20. Define: Pole of a mirror.
    

    ANS:The centre of the reflecting surface is called pole of the mirror.

    
    
21. What is the Principal axis of mirror?
    

    ANS:The imaginary line passing through the pole and the centre of curvature is called the principal axis of mirror.

    
    
22. Define: Aperture of mirror.
    

    ANS:The diameter of the reflecting surface is called aperture of mirror.

    
    
23. Define:Principal focus of a mirror.
    

    ANS:The point where the rays parallel to the principal axis meet or appear to meet after reflection is called the principal focus of the mirror.

    
    
24. Define:The focal length of mirror.
    

    ANS:The distance between the pole and the principal focus of a mirror is called the focal length of mirror.

    
    
25. What is the position, nature and size of the image formed by a concave mirror when the object is at infinity?
    

    ANS:The image is formed at focus; it is real, inverted and highly diminished.

    
    
26. What is the position, nature and size of the image formed by a concave mirror when the object is beyond the center of curvature of the mirror?
    

    ANS:The image is formed between the focus and the center of curvature; it is real, inverted and diminished.

    
    
27. What is the position, nature and size of the image formed by a concave mirror when the object is at C, the center of curvature of the mirror?
    

    ANS:The image is formed at the center of curvature; it is real, inverted and of the same size as the object.

    
    
28. What is the position, nature and size of the image formed by a concave mirror when the object is between C and F?
    

    ANS:The image is formed beyond C; it is real, inverted and enlarged.

    
    
29. What is the position, nature and size of the image formed by a concave mirror when the object is at F?
    

    ANS:The image is formed at infinite distance; it is real, inverted and highly enlarged.

    
    
30. What is the position, nature and size of the image formed by a concave mirror when the object is between pole and F?
    

    ANS:The image is formed behind the mirror; it is virtual, erect and enlarged.

    
    
31. Define: Magnification.
    

    ANS:The ratio of the height of the image to the height of the object is called magnification.

    
    
32. Define: Refraction.
    

    ANS:The change in the direction of a ray of light when it enters from one transparent medium to another transparent medium is called refraction.

    
    
33. Define: Optical center of a lens.
    

    ANS:Optical center of a lens is the center of lens on the principal axis.

    
    
34. Define: Radius of curvature of lens.
    

    ANS:The radii of spheres whose parts form the lens surfaces are called radii of curvature of lens.

    
    
35. Define: Principal focus of convex lens.
    

    ANS:The point of convergence of the rays parallel to principal axis on the principal axis is called the principal focus of a convex lens.

    
    
36. Define: Focal length of lens.
    

    ANS:The distance between the optical center and the principal focus is called the focal length of lens.

    
    
37. Mention the position, nature and size of the image formed by a convex lens when the object is placed at infinity.
    

    ANS:The image is on the opposite side of the object at focus; it is real, inverted and highly diminished(point sized).

    
    
38. Mention the position, nature and size of the image formed by a convex lens when the object is placed beyond 2F₁.
    

    ANS:The image is between F₂ and 2F₂; it is real, inverted and diminished.

    
    
39. Mention the position, nature and size of the image formed by a convex lens when the object is placed at 2F₁.
    

    ANS:The image is at 2F₂; it is real, inverted and of the same size as the object.

    
    
40. Mention the position, nature and size of the image formed by a convex lens when the object is between F₁ and 2F₁.
    

    ANS:The image is beyond 2F₂; it is real, inverted and enlarged.

    
    
41. Mention the position, nature and size of the image formed by a convex lens when the object is placed at focus F₁.
    

    ANS:The image is at infinite distance; it is real, inverted and highly enlarged.

    
    
42. Mention the position, nature and size of the image formed by a convex lens when the object is placed between focus F₁ and optical center O.
    

    ANS:The image is on the same side of the lens as the object; it is virtual, erect and enlarged.

    
    
43. Define: Power of lens.
    

    ANS:The reciprocal of the focal length(f) of a lens is called the power of that lens.

    
    
44. What is the SI unit of power of lens?
    

    ANS:The SI unit of power of lens is dioptre.

    
    
45. The power of a convex lens is positive.True or false?
    

    ANS:Yes, this is a true statement.

    
    
46. The power of a concave lens is positive.True or false?
    

    ANS:No, this is a false statement.

    
    
47. Which instrument is used to measure the power of a lens?
    

    ANS:Dioptremeter is used to measure the power of a lens.

    
    
48. What is the power of a convex lense whose focal length is 50 cm ?
    

    ANS:(+)2.0 D
    

    
    
49. The power of a lens is (-)4.0 D. What is its focal length in cm ?
    

    ANS:(-)25 cm, the negative sign indicates that the lens is concave.
    

    
    
50. On what does the lateral shift of an emergent ray depend in case of a rectangular glass slab ?
    

    ANS:The lateral shift is proportional to the perpendicular distance between two refracting surfaces of the glass slab.
    

    
    
51. Define: Critical angle.
    

    ANS:The angle of incidence of a ray for which the angle of refraction is 90° is called critical angle.

    
    
52. Define: Total internal reflection.
    

    ANS:When the angle of incidence of a ray is greater than the critical angle, the refracted ray is in the same medium, i.e. the light is totally reflected in the same medium. This is known as total internal reflection.

    
    
53. What happens to a ray of light travelling from optically rarer to optically denser medium?
    

    ANS:The ray of light bends towards the normal.

    
    
54. State Snell's law.
    

    ANS:The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.

    
    
55. Which type of waves are light waves?
    

    ANS:Light waves are electromagnetic(non-mechanical)waves.

    
    
56. A medium is necessary for the propagation of light waves.True or false?
    

    ANS:No, this is a false statement.

    
    
57. If incident angle is 40° for a plane mirror, what should be the angle of reflection?
    

    ANS:The angle of reflection should be 40°.

    
    
58. What is the type of mirror having focal length +10 cm?
    

    ANS:The mirror is convex.

    
    
59. Which medium has higher refractive index, glass or water?
    

    ANS:Glass has higher refractive index.

    
    
60. In which medium is the velocity of light higher, glass or water?
    

    ANS:The velocity of light is higher in water.

    
    
61. Give the names of two phenomena occuring due to refraction.
    

    ANS:Twinkling of stars and mirage are the two phenomena occuring due to refraction.

    
    
62. If the focal length of convex lens is 25 cm, what is its power?
    

    ANS:The power of the given convex lens is 4.0 D or +4.0 D.(HINT: p = 1/f where f is in meter)

    
    
63. Which kind of lens has positive power, converging or diverging ?
    

    ANS:A converging lens has positive power.

    
    
64. Mention the unit of refractive index.
    

    ANS:Refractive index is unitless.

    
    
65. Which type of mirror is used as rear-view mirror in vehicles ?
    

    ANS:A convex mirror is used as rear-view mirror in vehicles.

    
    
66. The refractive index of diamond is 2.42. What is the meaning of this statement in relation to the speed of the light ?
    

    ANS:It means that the light travels 2.42 times faster in vacuum than in diamond.

    
    

INTRODUCTION TO NANOTECHNOLOGY-III : LONG ANSWERS

GIVE ANSWER IN DETAIL

*INDEX-TOPIC SEARCH

1. EXPLAIN WITH DIAGRAM THE CONCEPT OF CARBON NANOTUBE.
   

   ANS:Richard Smalley conceptualised bucky balls, big enough to make tubes. It was made possible by bonding the bucky balls together in a row and truncating the overlapping regions of bucky balls.
   
   
   
   
   
   Nanotubes are cylinders in the form of lattice of carbon atoms.Each carbon atom is covalently bonded to three other carbon atoms. Hence carbon nanotubes are basically a chain of bucky balls whose ends never close into a sphere when they are formed and a lattice of atoms in the shape of a cylinder is formed.
   
   
   
   
   Nanotubes can be Single-Walled Nano Tube(SWNT) or Multi-Walled Nano Tube(MWNT).SWNT consists of a single cylinder whereas MWNT consists of multiple concentric nanotubes of different diameters.
   

   
   
2. DISCUSS THE CHARACTERISTICS AND PROPERTIES OF CARBON NANOTUBES.
   

   ANS:The characteristics and properties of carbon nanotubes are:
   

   1. Tensile Strength:
      
      • The tensile strength of a carbon nanotube is about 100 times more than that of steel.
        
      • High tensile strength is due to the carbon-carbon bonds and the fact that each carbon nanotube is one large molecule.
        
      • Carbon nanotubes are elastic despite their high tensile strength.Therefore they can be bent like a rubber tube.
      
      
   2. Thermal Conductivity:
      
      • The thermal conductivity of carbon nanotubes is 10 times that of silver.
        
      • Carbon nanotubes conduct heat by vibrations of the covalent bonds between the carbon atoms. The atoms wiggle around themselves and transmit heat through the material.
        
      • Because the bonds in the molecule are elastic like spring the vibrations occur.
        
      • These vibrations transmit quickly through the tube due to the stiffness of the tube.
      
      

   
   

   
   
3. WHAT IMPROVEMENT WILL NANOTECHNOLOGY BRING ABOUT? EXPLAIN.
   

   ANS:The nanotechnology is expected to bring about the following improvement in near future:
   

   • Global Positioning System(GPS) will be more accurate, smaller and cost-effective.
     
   • Computers will be smaller and faster.
     
   • Super-hard materials with tunable melting points will be made.
     
   • Cell-phone batteries will have longer life.
     
   • DNA fingerprinting will become quicker and more accurate.
     
   • Medical diagnosis and delivery systems will be more efficient.
     
   • Active nanoingredients in sunscreen creams will bring quick skin health.
     
   • Information routing will be done at the speed of light.

   
   

   
   
4. DESCRIBE THE IMPORTANCE OF CARBON IN NANOTECHNOLOGY.
   

   ANS:The importance of carbon in nanotechnology is as follows:
   

   • Carbon is found in great abundance in natural materials.
     
   • Materials containing carbon exhibit a wide range of properties due to specific reasons.
     
   • A carbon atom can form covalent bonds with different types of atoms.
     
   • A carbon atom can form covalent bonds with four other atoms at a time resulting in long-chained molecules.
     
   • A carbon atom can bond strongly to other carbon atoms in many different ways forming a variety of structures.
     
   • All this makes carbon an important element in nanotechnology where the basic structure is a Bucky Ball which is a molecule of 60 carbon atoms forming a sphere-like structure.
     
   • A series of such spheres (Bucky Balls) connected to one another without the ends closing form carbon nanotubes which have many applications in nanotechnology.
     
   • All this makes carbon a very important element for nanotechnology.

   
   
5. EXPLAIN HOW NANOTECHNOLOGY PLAYS AN IMPORTANT ROLE IN SECURITY.
   

   ANS:Nanotechnology plays an important role in security in the following way:
   

   • Superior and Lightweight Material:Nanomaterials ten times stronger than steel will revolutionise battle tanks, spacecrafts, skyscrappers and bridges. We can tune the melting points of nanomaterials by controlling their particle size in the range of nanoscale.
     
   • Powerful Munitions:New nanometals and nanosize particles such as nanoaluminium are chemically more reactive. Varying the size of these particles in munitions will cause minimum collateral damage during explosion.
     
   • Advanced Computing:More powerful and small computers will encrypt date and provide round-the-clock security,e.g. Quantum Cryptography.
     
   • Powerful Chemical Sensors:We can develop highly sensitive chemical sensors which can accurately pinpoint a single molecule out of billion molecules floating around.These sensors will be cheaper and disposible.
     
   
   

   
   

   
   
6. WRITE A NOTE ON NANOTECHNOLOGY AND HEALTHCARE.
   

   ANS:Nanotechnology has wide scope of application in the field of healthcare.
   

   • Diagnostics:
     
     1. Better,cheaper and quicker diagnostic equipment will enable instant diagnosis and drug application.
        
     2. For example:Floating contrast agents into bloodstream will allow detection of diseases with accuracy and speed.
        
     3. Quick mapping of DNA for newly-born babies will give information of potential problems likely to occur in future.This will enable us to curtail diseases at the early stage of life.
     
     
     
     
   • Novel Drugs:Nanotechnology can make drug delivery in precise amount and to precise site in our body.For example: A nanoshell of 100 nm diameter will float through the body, attaching only to cancerous cells.Upon excitation by a laser beam those nanoshells will dissipate heat and the tumour will get destroyed.
     
   
   

   
   

   
   
7. MENTION SOME BENEFITS OF NANOTECHNOLOGY TO MANKIND.
   

   ANS:Some benefits of nanotechnology to mankind are:
   

   • Nanoparticle medicines with vastly improved delivery and control.
     
   • Highly improved printing by nanoscale particles with best properties of both dyes and pigments.
     
   • Vastly improved lasers and magnetic disc heads made by controlling layer thickness.
     
   • 'Nanofood' molecularly identical to the organic food.
     
   • Better, cheaper and quicker diagnostic equipment to enable instant diagnosis and drug application.
     
   • Drug delivery in precise amount to precise site in our body for the treatment of fatal diseases like cancer.
     
   • Better,efficient and cheaper means of security with the help of superior and lightweight materials, powerful munitions, advanced computing and powerful chemical sensors.
     
   • Transformation of electricity into light with the help of nanocrystals preventing excessive loss of energy due to heating.
     
   • Inexpensive and more efficient solar cells and hydrogen fuel cells.
     
   • Lighter, stronger and more efficient engine blocks for cars to save fuel.
     
   • Efficient water purification technology and thus access to clean water in third world countries.
     
   
   

   
   

   
   

INTRODUCTION TO NANOTECHNOLOGY-II : BRIEF ANSWERS

ANSWER IN BRIEF

*INDEX-TOPIC SEARCH

1. DEFINE: NANOSCIENCE AND NANOTECHNOLOGY IN THE MODERN CONTEXT.
   

   ANS:Nanoscience is the study of fundamental principles of molecules and other systems whose at least one of the dimensions lies between 1 nm and 100 nm range.
   
   Nanotechnology is the tool to harness the knowledge of nanoscience for creating new useful materials, devices, machines and systems with novel properties due to their small size. Nanotechnology is building structures, atom-by-atom or molecule-by-molecule, that will be helpful in manufacturing devices and systems.
   
   Nanoscience and nanotechnology will find the ways to make stronger materials,detect diseases in blood-stream, build tiny machines, generate light and enormous energy and purify water with the help of nanosize particles.
   

   
   
2. WHY IS NANOTECHNOLOGY A GENERAL-PURPOSE TECHNOLOGY ?
   

   ANS:Nanotechnology is a general purpose technology because in its mature form it will have significant impact on almost all industries and all areas of society. It offers better built, longer lasting, cleaner, safer and smarter products useful in domestic purposes, communication, medicines, transportation, agriculture and industry.
   

   
   
3. WHY IS CARBON A FUNDAMENTAL ELEMENT IN NANOTECHNOLOGY? EXPLAIN.
   

   ANS:Carbon is a fundamental element in nanotechnology due to the following properties.
   

   1. A carbon atom can bond with many different types of atoms by forming covalent bonds.
      
   2. Each carbon atom can form such covalent bonds with four other atoms at a time and thus form long chains of atoms.
      
   3. No other element except carbon can bond strongly to other carbon atoms in many different ways.
      

   
   

   
   


4. EXPLAIN HOW NANOTECHNOLOGY IS USED IN DIAGNOSTICS.
   

   ANS:Nanotechnology is useful in diagnostics in the following way:
   

   1. Better, cheaper and quicker diagnostic equipment will enable instant diagnosis and drug application.
      
   2. For example, floating contrast agents into bloodstream will allow detection of disease with accuracy and speed.
      
   3. Quick mapping of DNA for newly born babies will give information of potential problems likely to occur in future.This will enable us to curtail diseases at early stage of life.
      

   
   

   
   


5. WHICH FIELDS OF TECHNOLOGY WILL BE IMPORTANT IN FUTURE? or MENTION SOME IMPORTANT FUTURE TECHNOLOGIES THAT WILL BE IN FOREFRONT DURING 21ST CENTURY.
   

   ANS:The fields of technology that will be important in future are:
   

   1. Nanotechnology
      
   2. Biotechnology
      
   3. Information Technology
      
   4. X-Rays
      
   5. Teleportation
      
   6. LASERS
      
   7. Robotics
      

   
   



6. WHAT POSSIBILITIES WILL NANOTECHNOLOGY OFFER ?
   

   ANS:Nanotechnology will offer following possibilities such as :
   

   
   
   1. Cellular repair
   2. Artificial intelligence
   3. Inexpensive space travel
   4. Clean and abundant energy.
   5. Environmental restoration.
      

   
   

   
   
7. EXPLAIN THE INFLUENCE OF NANOTECHNOLOGY ON ENVIRONMENT.
   

   ANS:Nanotechnology will have great influence on environment.
   

   1. It would help repair past environmental damages and thus reform the environment.
      
   2. It can solve present environmental problems.
      
   3. It would prevent future environmental impacts.
      
   4. It would sustain the planet Earth for future generations.
      

   
   



8. DESCRIBE THE STRUCTURE AND CONFIGURATION OF A BUCKY BALL.
   

   ANS:
   
   
   A bucky ball is a molecule of 60 carbon atoms(C₆₀) in the architectural configuration of a soccer ball(sphere).Each carbon atom is bound to three adjacent carbon atoms forming a sphere of around 1.0 nm diameter.The bonds between atoms form a pattern of joined hexagons and pentagons similar to the panels on a soccer ball.
   



9. GIVE THE APPLICATIONS OF BUCKY BALLS AND CARBON NANOTUBES.
   

   ANS:The applications of bucky balls and carbon nanotubes(i.e.,nanotechnology) are:
   

   1. In electronics
   2. In food and agriculture
   3. In sports and toys
      
   4. In spintronics
   5. In photonics
   6. In display panels
   7. In space vehicles
      
   8. In the field of energy
   9. In medicine and diagnosis
   10. In consumer goods
       

   
   



10. "NANOTECHNOLOGY PLAYS BY DIFFERENT RULES." EXPLAIN THE STATEMENT.
    

    ANS:How matter behaves at nanoscale is the main aspect of nanotechnological research. The rules that govern large systems may not work at this scale. This is because nanomaterials have larger surface area-to-volume ratio.Therefore, phenomena like sticking and friction have much more dominance and importance at nanoscales as compared to large dimensions.Thus nanotechnology plays by different rules.
    



11. MENTION THE USE OF NANOTECHNOLOGY FOR RESOURCES LIKE ENERGY AND WATER.
    

    ANS:

    
    
    1. ENERGY : Nanocrystals will transform electricity into light without loss of energy due to heating. Solar cells and hydrogen fuel cells will become inexpensive and efficient. Cars will become more efficient as lighter and stronger engine blocks will save fuel.
       
    2. WATER :Efficient water purification with the help of nanotechnology will allow third-world countries to have access to clean water.
       

    
    

    
    
12. WHAT CAN WE EXPECT TO BE IMPROVED IN NEAR FUTURE DUE TO NANOTECHNOLOGY ?
    

    ANS:

    
    
    • Smaller and cost-effective accurate Global Positioning System(GPS)
    • Faster and smaller computers
    • Super hard materials with tunable melting temperatures
    • Cell phones with longer battery life
    • Quick and accurate DNA finger-printing
    • New and efficient medical diagnosis and delivery systems
    • Quick skin healing by sun screen creams having active nano-ingredients
    • Information routing at the speed of light.
      

    
    

    
    
13. WHAT CAN WE DO WITH THE HELP OF STUDY OF NANOSIZE PARTICLES, DEVICES AND COMPOSITES ?
    

    ANS:With the help of the study of nanosize particles, devices and composites we can develop stronger materials, detect diseases in blood streams, build tiny machines, generate light and enormous energy and purify water.
    

    
    
14. WRITE A NOTE ABOUT AFM AND STM.
    

    ANS: AFM(Atomic Force Microscope) and STM(Scanning Tunelling Microscope) are instruments used to see objects of size smaller than 10000 nm since the human eye has the limitation that it cannot see objects smaller than 10000 nm.
    Gern Binnig and Heinrich Rohre of IBM Research Lab invented STM in 1981.
    With the help of such instruments(machines) we can study atoms and manipulate them to develop new nanostructures.