## Light Refraction Solved Examples

**Ex.1 **Speed of light in water is 2.25 × 108 m/s. Calculate the refractive index of water.

**Sol. ** Refractive index is given by

n =

= = 1.33

**Ex.2 **Refractive index of diamond is 2.42. Calculate the speed of light in diamond.

**Sol. ** We know that refractive index,

n =

or 2.42 =

or n == 1.24 × 10^{8} m/s.

** ****Ex.3 **A ray of light travelling in air falls on the surface of water. The angle of incidence is 60° with the normal to the surface. The refractive index of water = 4/3. Calculate the angle of refraction.

**Sol. **

We know that = n

Here, i = 60°, n = 4/3

∴

or

or sin r = = 0.65

∴ r = sin^{–1} 0.65

** **

**Ex.4 **If the refractive index of water is 4/3 and that of glass is 3/2. Calculate the refractive index of glass with respect to water.

**Sol. ** We known that

** **^{w}m_{g }=

where ^{w}m_{g }= refractive index of glass with

respect to water

m_{g }= refractive index of glass = 3/2

m_{w }= refractive index of water = 4/3

\ ^{w}m_{g }= = =1.1

**Ex.5 **A ray of light is incident on the plane surface of a transparent medium at an angle 60° with the normal. The angle of refraction is 30°. Calculate the refractive index of the transparent material.

**Sol. ** Here,

Angle of incidence, i = 60°

Angle of refraction, r = 30°

Refractive index,

n = = = =

**Ex.6 **A ray of light travelling in air falls on the surface of a glass slab at an angle 45° with the normal. The refractive index of glass is 1.5. Calculate the angle of refraction.

**Sol. ** Angle of incidence = 45°

Refractive index of glass, n = 1.5

Since n =

or 1.5 =

or sin r = =

=

= = = 0.47

\ r = sin^{–1}0.47

**Ex.7 **The refractive index of diamond is 2.42 and that of carbon disulphide is 1.63. Calculate the refractive index of diamond with respect to carbon disulphide.

**Sol. ** Refractive index of carbon disulphide,

n1= 1.63

Refractive index of diamond, n2 = 2.42

\ Refractive index of diamond with respect for carbon disulphide,

^{1}n_{2} = = = 1.48

**Ex.8 **A coin is placed in a tumbler, water is then filled in the tumbler to a height of 20 cm. If the refractive index of water is 4/3, calculate the apparent depth of the coin.

**Sol. ** Here,

Real depth, h = 20 cm

Refractive index, n = 4/3

Now, n =

or =

or Apparent depth = = 15 cm

**Ex.9 **There is a black spot on a table. A glass slab of thickness 6 cm is placed on the table over the spot. Refractive index of glass is 3/2. At what depth from the upper surface will the spot appear when viewed from above?

**Sol. ** Real depth of the spot = 6 cm

Refractive index of glass, n =

Now, n =

or =

\ Apparent depth = = 4 cm

** **

**Ex.10 **Refractive index of diamond is 2.42 and that of glass is 1.5. Calculate the critical angle for diamond-glass surface.

**Sol. ** Refractive index of diamond, n_{1} = 2.42

Refractive index of glass, n_{2} = 1.5

Now, sin ic =

= = 0.6198

\ ic = sin^{–1 }0.62

**Ex.11 **Refractive index of glass is 3/2. A ray of light travelling in glass is incident on glass-water surface at an angle 30° with normal. Will it be able to come out into the water Refractive index of water = 4/3.

**Sol. ** Refractive index of glass, n1 = 3/2

Refractive index of water, n2 = 4/3

Now, sin ic = = = = 0.88

\ ic = 62°

Since, the angle of incidence (30°) is less than the critical angle, the ray will be refracted into the water.

**Ex.12 **The refractive index of dense flint glass is 1.65 and that of alcohol is 1.36, both with respect to air. What is the refractive index of flint glass with respect to alcohol ?

**Sol. ** Refractive index of flint glass, n_{2} = 1.65

Refractive index of alcohol, n_{1} = 1.36

\ Refractive index of flint glass with respect to alcohol is given by

^{1}n_{2} = = = 1.21

** **

**Ex.13 **An object is placed 36 cm from a convex lens. A real image is formed 24 cm from the lens. Calculate the focal length of the lens.

**Sol. ** According to the sign convention the object is placed on the left-hand side of the lens. So object distance (u) is negative. Real image is formed on the other side of the lens. So the image distance (n) is positive. Thus, u = –36 cm, n = +24 cm, *f* = ?

Using lens formula, – = , we get

– =

or = + =

\ *f *= = 14.4 cm

** **

**Ex.14 **A 2 cm long pin is placed perpendicular to the principal axis of a lens of focal length 15 cm at distance of 25 cm from the lens. Find the position of image and its size.

**Sol. ** Here, u = –25 cm, * f* = +15

Using the lens formula, – = we get

or – =

or = – =

or n = = 37.5 cm

The positive sign shows that the image is formed on the right-hand side of the lens.

Magnification is given by

m = =

or = = = – 1.5

\ h = – 1.5 × h = –1.5 × 2 cm

= –3 cm

The image of the pin is 3 cm long. The negative sign shows that it is formed below the principal axis, i.e. the image is inverted.

**Ex.15 **A point object is placed at a distance of 18 cm from a convex lens on its principal axis. Its image is formed on the other side of the lens at 27 cm. Calculate the focal length of the lens.

**Sol. ** According to the sign convention, the object is placed on the left-hand side of the lens, therefore object-distance is negative,

i.e. u = –18 cm. Since the image is formed on the other side, the image-distance is positive, i.e., v = +27 cm. Using lens formula,

–=, we have

– =

or + = =

or f = = 10.8 cm

**Ex.16 **A convex lens forms an image of the same size as the object at a distance of 30 cm from the lens. Find the focal length of the lens. Also find power of the lens. What is the distance of the object from the lens ?

**Sol. ** A convex lens forms the image of the same size as the object only when the object is placed at a distance 2f from the lens. In this case the image is also equal to 2f from the lens.

Hence, 2f = 30 cm

or f = 15 cm = 0.15 m

Power of the lens, P = = D = 6.6D

The distance of the object from the lens is also 2f = 30 cm.

** **

**Ex.17 **A 3 cm high object is placed at a distance of 80 cm from a concave lens of focal length 20 cm. Find the position and size of the image.

**Sol. ** Here, u = –80 cm, f = – 20 cm

Using the lens formula, – = , we get

or – =

or = – – == –

or n = –16 cm

Magnification, m = = = =

or h’ == = 0.6 cm

Length of image is 0.6 cm. Positive sign shows that the image is erect.

**Ex.18 **An object is placed on the principal axis of a concave lens at a distance of 40 cm from it. If the focal length of the lens is also 40 cm, find the location of the image and the magnification.

**Sol. ** For a concave lens focal length f is negative, i.e. f = –40 cm. Since by convention, object is placed on the left of the lens, so u = – 40 cm.

Using the lens formula, – = , we get

or – =

or = – – = –

or n = – 20 cm

The image is formed 20 cm from the lens. Minus sign shows that the image is formed on the same side of the lens as the object.

Now, magnification, m = = = =

Positive sign shows that the image is erect.

**Ex.19 **A beam of light travelling parallel to the principal axis of a concave lens appears to diverge from a point 25 cm behind the lens after refraction. Calculate the power of the lens.

**Sol. ** When a parallel beam after refraction through the lens is incident on a concave lens, it appears to diverge from the focus of the lens. Hence, the focal length of the lens is 25 cm. According to sign convention, focal length of a concave lens is negative.

\ f = –25 cm = –0.25 m

\ Power, P = = = – 4D

**Ex.20 **A convex lens of power 5D is placed at a distance of 30 cm from a screen. At what distance from the lens should the screen be placed so that its image is formed on the screen?

**Sol. ** Power of the lens, P = +5D

\Focal length, f = = = 0.20 m = 20 cm

Here, the screen is placed 30 cm from the lens.

\ v = +30 cm, f = +20 cm, u = ?

Using the lens formula, – = , we get

– =

or = – = –

or u = – 60 cm

Therefore, the screen should be placed at 60 cm from the lens.

**Ex.21 **A pin 3 cm long is placed at a distance of

24 cm from a convex lens of focal length 18 cm. The pin is placed perpendicular to the principal axis. Find the position, size and nature of the image.

**Sol. ** Here, u = –24 cm, f = +18 cm, v = ?

Using the lens formula, – = , we get

– =

or = – =

or v = 72 cm

The image is formed 72 cm from the lens on the other side. So the image is real.

Magnification, m = = = = –3

or h’ = –3 × h = –3 × 3.0 = – 9 cm

The image is 9 cm in size. Negative sign shows that the image is inverted.

**Ex.22 **A convex lens of focal length 40 cm and a concave lens of focal length 25 cm are placed in contact in such a way that they have the common principal axis. Find the power of the combination.

**Sol. ** Focal length of the convex lens,

** **f_{1} = 40 cm = +0.4 m

\ Power of the convex lens,

P1= = +2.5D

Focal length of the concave lens,

f_{2} = –25 cm = –0.25 m

\ Power of the concave lens,

P_{2} = = – 4D

Power of the combination,

P = P_{1} + P_{2} = 2.5 – 4D = – 1.5D

**Ex.23 **A concave lens has a focal length of 15 cm. At what distance should the object be from the lens placed so that it forms an image 10 cm from the lens ? Also find the magnification.

**Sol. ** A concave lens always forms a virtual, erect image on the same side as the object.

Image distance, v = –10 cm

Focal length f = –15 cm

Object distance, u = ?

Using, the lens formula, –=, we get

or – =

or – = = –

or u = –30 cm

Thus, the object should be placed 30 cm on the lens.

Magnification, m = = = = 0.33

The positive sign shows that the image is erect and virtual. The size of the image is one-third of that of the object.

**Ex.24 **A 2 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of object from the lens is 15 cm. Find the position, nature and size of the image. Calculate the magnification of the lens.

**Sol. ** Object distance, u = –15 cm

Focal length, f = +10 cm

Object height, h = +2 cm

Image distance, v = ?

Image height, h’ = ?

Using the lens formula, – = , we get

= =

or = – =

or n = +30 cm

Positive sign of v shows that the image is formed at a distance of 30 cm on the right side of the lens. Therefore the image is real and inverted

Magnification, m = =

= = –2

or h’ = –2 × 2 = –4 cm

Magnification, m = = = –2

Negative sign with the magnification and height of the image shows that the image is inverted and real. Thus, a real image of height 4 cm is formed at a distance of 30 cm on the right side of the lens. Image is inverted and twice the size of the object.