Question

The convex surface of a thin concavo-convex lens of glass of refractive index 1.5 has a radius of curvature 20 cm. The concave surface has a radius of curvature 60 cm. The convex side is silvered and placed on a horizontal surface as shown in figure. (a) Where should a pin be placed on the axis so that its image is formed at the same place? (b) If the concave part is filled with water (μ = 4/3), find the distance through which the pin should be moved so that the image of the pin again coincides with the pin.
Figure

Answer 1

Object must be placed at 15 cm from the lens on the axis.

Explanation:

(a) Let the pin be x far from the lens

Then, the refraction is ,

$\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}$

Here,

$\mu_{2}=1.5$

$\mu_{1}=1$

u = -x

R = -60 cm  

$\frac{1.5}{v}-\frac{1}{-x}=\frac{1.5-1}{-60}$

$\frac{1.5}{v}+\frac{1}{x}=-\frac{0.5}{60}$

$\frac{1.5}{v}+\frac{1}{x}=-\frac{5}{600}$

$\frac{1.5}{v}+\frac{1}{x}=-\frac{1}{120}$

$1.5 x+v=-\frac{1}{120}$

120(1.5x+v) = - vx   -----> eqn (1)

180x + 120v = - vx

120v + vx = -180x

v(120+x) = -180x

$v=\frac{-180 x}{(120+x)}$        

This picture distance for the concave mirror is again object distance.

$u=\frac{-180 x}{(120+x)}$

$f = -10 cm$

$f=\frac{R}{2}$

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

$\frac{1}{\frac{-180 x}{(120+x)}}+\frac{1}{v}=\frac{1}{-10}$

$\frac{1}{v_{1}}=-\frac{1}{10}-\frac{1}{\frac{-180 x}{(120+x)}}$

$\frac{1}{v_{1}}=-\frac{1}{10}+\frac{(120+x)}{180 x}$

$\frac{1}{v_{1}}=\frac{-18 x+(120+x)}{180 x}$

$\frac{1}{v_{1}}=\frac{120+x-18 x}{180 x}$

$\frac{1}{v_{1}}=\frac{120-17 x}{180 x}$

$v_{1}=\frac{180 x}{120-17 x}$

Once, the image created through the lens is refracted so that the image is produced on the object taken in the 1st refraction. So, for refraction 2nd.

Conversion according to sign

v = -x      

$\mu_{2}=1$

$\mu_{1}=1.5$

R = -60

$\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R}$

$\frac{1}{-x}-\frac{1.5}{u}=\frac{1-1.5}{-60}$

$\frac{1}{-x}-\frac{1.5}{u}=\frac{-0.5}{-60}$

$\frac{1}{-x}-\frac{1.5}{u}=\frac{0.5}{60}$

$-\frac{1}{x}-\frac{1.5}{u}=\frac{5}{600}$

$-\frac{1}{x}-\frac{1.5}{\frac{180 x}{120-17 x}}=\frac{1}{120}$

$\frac{1}{x}+\frac{120-17 x}{120 x}=\frac{-1}{120}$

Multiply the two sides by 120

We get,    

120+120-17x=-x

120+120=-x+17x

16x=240

$x=\frac{240}{16}$

x = 15 m

Thus, the target should be positioned on the axis at 15 cm away from the lens .