*INDEX-TOPIC SEARCH

**If an object of height 4 cm is placed at distance of 12 cm from a concave mirror having focal length 24 cm, find the position, nature and the height of the image.****SOLUTION**

Here,

object height h = 4 cm

object dis tance u = -12 cm

focal length f = -24 cm

Putting these values in mirror formula,

(1/u) + (1/v) = (1/f)

∴ (-1/12) + (1/v) = (-1/24)

∴ (1/v) = (1/12) - (1/24)

∴ (1/v) = (1/24)

∴ v = 24

Now, magnification m = (-u/v) = (-24)/(-12) = 2

∴ (h'/h) = 2

∴ h' = 2h = 2 x 4 = 8 cm

Since v is positive, the image is formed behind the mirror at the distance of**24 cm**from the mirror.It is virtual and has height of**8 cm**.**An object of height 6 cm is placed at a distance of 10 cm from a convex mirror with radius of curvature 30 cm.Find the position, nature and the height of its image.****SOLUTION**

Here,Object distance u = -10 cm

Radius of curvature R = 30 cm

Object height h = 6 cm

Now,

focal length f = (R/2) = 30/2 = +15

Using mirror formula,

(1/u) + (1/v) = (1/f)

∴ (1/v) = (1/f) - (1/u)

∴ (1/v) = (1/15) - (1/-10)

∴ (1/v) = (1/15) + (1/10) =(1/6)

∴ v = 6 cm

Also, magnification m = -(u/v) = -(6/-10) = 0.6

∴ (h'/h) = 0.6

∴ h' = mh =0.6 x 6 = 3.6 cm

Thus, the image is formed behind the mirror(as v is positive), it is virtual and its height is**3.6**cm.**The refractive index of water with respect to air is 1.33. The velocity of light in vacuum is 3 x 10**^{8}m/s. Calculate the velocity of light in water.**SOLUTION**

Here, the refractive index of water with respect to air is 1.33.

∴ absolute refractive index of water = 1.33

Now, Absolute refractive index of water =__Velocity of light in vacuum__

Velocity of light in water

∴ Velocity of light in water =__Velocity of light in vacuum__

Absolute refractive index of water

=__3 x 10__^{8}

1.33

= 2.25 x 10^{8}m/s.**A ray of light enters from water to glass.Refractive index of glass with respect to water is 1.12. Find absolute refractive index of water if absolute refractive index of glass is 1.5.****SOLUTION**

Taking water as "medium1" and glass as "medium2",

Absolute refractive index of water, η_{1}= ?

Absolute refractive index of glass, η_{2}= 1.5

η_{21}= 1.12_________(given)

∴ η_{2}/η_{1}= 1.12

∴ η_{1}=η_{2}/1.12

∴ η_{1}= (1.5)/(1.12) = 1.34

Hence,absolute refractive index of water = 1.34**A swimmer lights a torch under sea water.Light from the torch is incident on water surface in such a way that incident light makes an angle of 37° with water surface.Find the angle of refraction if absolute refractive indices of water and air are 1.33 and 1.0 respectively.****SOLUTION**

Taking water as "medium1" and air as "medium2",we have

η_{1}= 1.33

η_{2}= 1.00

Angle of incidence, θ_{1}= 37°

Angle of refraction, θ_{2}= ?

Now, (sin θ_{1}/sin θ_{2}) = (η_{2}/η_{1})

∴ (sin 37°/sin θ_{2}) = (1.00/1.33)

∴ sin θ_{2}= [ (0.6015 x 1.33)/1.00]

∴ sin θ_{2}= 0.8000

∴ θ_{2}= 53°_____(from the table of sines)

Hence, angle of refraction = 53°**An object is placed at 30 cm in front of a convex lens of focal length 20 cm. Find the position of the image.****SOLUTION**

Here,

Focal length f = +20 cm

Object distance u = -30 cm

Image distance v = ?

Using lens formula,

(1/f) = (1/v) - (1/u)

∴ (1/v) = (1/u) + (1/f)

∴ (1/v) = (1/-30) + (1/20)

∴ (1/v) = (-2 + 3)/60

∴ (1/v) = (1/60)

∴ v = 60 cm.

The positive value of v indicates that the image is formed at 60 cm on the right side of lens from the center of the lens.**At what distance the object should be placed so that the image will be formed at a distance 10 cm from a concave lens ? Focal length of the lens is 20 cm.****SOLUTION**

Since the lens is concave, the image will be formed on the same side as that of the object.

Here,

Image distance v = -10 cm

Focal length f = -20 cm

Object distance u = ?

Using lens formula,

(1/f) = (1/v) - (1/u)

∴ (1/u) = (1/v) - (1/f)

∴ (1/u) = (1/-10) - (1/-20)

∴ (1/u) = (-2 + 1)/20

∴ (1/u) = -(1/20)

∴ u = -20 cm

Thus the object should be placed 20 cm from the concave lens on the left side of the lens.

**An object of 5 cm height is placed at a distance of 15 cm from a concave mirror.Find the position, height and nature of its image.The focal length of the mirror is 10 cm.****SOLUTION**

Here

height of object, h = 5 cm

Distance of object, u = -15 cm

Focal length of concave mirror, f = -10 cm

Distance of image, v = ?

Using mirror formula,

(1/v) + (1/u) = (1/f)

∴ (1/v) = (1/f) - (1/u)

∴ (1/v) = (1/-10) - (1/-15)

∴ (1/v) = -(1/10) + (1/15)

∴ (1/v) = (-3 + 2)/30 = -(1/30)

∴ v = -30 cm

Also,(image height)/(object height) = (-v/u)

∴ (h'/h) = (-v/u)

∴ (h'/5) = (-30/-15)

∴ h' = 10 cm

Hence the image is formed at a distance of 30 cm ; its height is 10 cm and it is real and inverted.**An object of 10 cm height is placed at a distance of 10 cm from a convex mirror.The radius of curvature of the mirror is 30 cm.Find the position, height and nature of its image.****SOLUTION**

Here

Height of object, h = 10 cm

Distance of object, u = -10 cm

Radius of curvature of convex mirror, R = 30 cm

Distance of image, v = ?

Since f = (R/2),

f = (30/2) = 15 cm

Using mirror formula,

(1/u) + (1/v) = (1/f)

∴ (1/v) = (1/f) - (1/u)

∴ (1/v) = (1/15) - (1/-10) = (1/15) + (1/10)

∴ (1/v) = (2 + 3)/30 = (1/6)

∴ v = 6 cm

Also,(image height)/(object height) = (-v/u)

∴ (h'/h) = (-v/u)

∴ (h'/10) = (-6/-10)

∴ h' = 6 cm

Hence the image is formed at a distance of 6 cm; its height is 6 cm and it is virtual and erect.**Rays of light are entering from glass to glycerin.If the absolute refractive index of glass is 1.5 and that of glycerin is 1.47 then find the refractive index of glycerin with respect to glass.****SOLUTION**

Let's consider glass as medium1 and glycerin as medium2.

Absolute refractive index of glass, η_{1}= 1.5

Absolute refractive index of glycerin, η_{2}= 1.47

Now,

Refractive index of glycerin with respect to glass = η_{21}

=η_{2}/η_{1}= 1.47/1.5 = 0.98

Hence the refractive index of glycerin with respect to glass is 0.98.**Rays of light are entering from air to water. If the angle of incidence at the surface separating two mediums is 70°, find the angle of refraction of light in water.Absolute refractive index of water is 1.33.****SOLUTION**

Let's consider air as medium1 and water as medium2.

Absolute refractive index of air, η_{1}= 1.00

Absolute refractive index of water, η_{2}= 1.33

Angle of incidence θ_{1}= 70°

Angle of refraction θ_{2}= ?

Now, sin θ_{1}/sin θ_{2}= η_{2}/η_{1}

∴ (sin 70°/sin θ_{2}) = (1.33/1.00)

∴ (0.9397/sin θ_{2}) = (1.33/1.00)

∴ sin θ_{2}= 0.7065

∴ θ_{2}= 44° 57'

Hence the angle of refraction = 44° 57'**An object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm at a distance of 20 cm from the lens. Find the position of the image.****SOLUTION**

Here

Focal length of convex lens, f = 10 cm

Distance of object, u = -20 cm

Distance of image, v = ?

Using lens formula,

(1/f) = (1/v) - (1/u)

∴ (1/v) = (1/f) + (1/u)

∴ (1/v) = (1/10) + (1/-20) = (1/10) - (1/20)

∴ (1/v) = (2 - 1)/20 = (1/20)

∴ v = 20 cm

Hence the image is formed at the distance of 20 cm from the lens.**An object is placed perpendicular to the principal axis of a concave lens of focal length 30 cm at a distance of 20 cm from the lens. Find the position of the image.****SOLUTION**

Here

Focal length of concave lens, f = -30 cm

Distance of the object, u = -20 cm

Distance of the image, v = ?

Using lens formula,

(1/f) = (1/v) - (1/u)

∴ (1/v) = (1/f) + (1/u)

∴ (1/v) = (1/-30) + (1/-20) = (-2 - 3)/60

∴ (1/v) =(-5/60) = (-1/12)

∴ v = (-12) cm

Hence the image is formed at the distance of 12 cm from the lens on the same side of the object.**Find the power of a convex lens with focal length 0.4 m****SOLUTION**

Power of lens, p = (1/f)

∴ p = (1/0.4) ______________(for convex lens f is +ve)

∴ p = 2.5 D

Hence the power of lens is 2.5 D (dioptre).

## 12 comments:

Thanx for these Numericals.....!!

It helped me alot...!!

thanx so much for the numericals it helped ne a lot in the completion of my holiday assignment

the problems are quite easy

try and make the sums much more difficult

anyway thanks a lot sir

thank you

thanks a lot........these questions helped me for pratice the numericals of light.............

a hearty thanks to u sir ....

i was searching 4 this typ of question paper n i got it through u !

thanks a lot again !!

Nice numericals helped me a lot

Outstanding approach but pls feed us some more

thank you....... your site helped me a lot. also, it helped me to take print out of the questions and answers.

keep your work going. :)

sums are good

good sums! they helped me lot! thank you soo much sir!

Post a Comment