LIGHT:REFLECTION AND REFRACTION-III : LONG ANSWERS

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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: