Question

The convex surface of a thin concavo– convex lens of glass of refractive index $ 1.5 $ has a radius of curvature $ 20cm $ . The concave surface has a radius of curvature of $ 60cm $ . The convex side is coated with silver and placed at a horizontal surface as shown in the figure.

(a) Where a pin should be placed on the optical axis such that its image is formed at the same place?
(b) If the concave part is filled with water of refractive index $ \dfrac{4}{3} $ , find the distance through which the pin should be moved, so that the image of the pin again coincides with the pin?

Answer 1

Hint: A lens whose one side is concave and the other side is convex is called concavo convex lens. There are three processes which take place in this setup. Firstly refraction takes place, then reflection and then it again refracts.

Complete step by step answer:
(a)
Let the pin be placed at $ x $ distance from the lens.
Let us assume that the refractive index of air is $ {\mu _1} $ and that of glass is $ {\mu _2} $
 $ {\mu _1} = 1 \\
  {\mu _2} = 1.5 \\ $
Let us assume that the radius of convex curvature is $ {R_1} $ and the radius of concave curvature is $ {R_2} $
 $ {R_1} = 20cm \\
  {R_2} = 60cm \\ $
When refraction takes place,
 $ \dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R} $
Substituting the values according to sign convention ,we get:
 $ \dfrac{{1.5}}{v} - \dfrac{1}{{ - x}} = \dfrac{{1.5 - 1}}{{ - 60}} \\
  \dfrac{{1.5}}{v} = \dfrac{{0.5}}{{ - 60}} + \dfrac{1}{{ - x}} \\
  \dfrac{{1.5}}{v} = \dfrac{{ - 0.5x - 60}}{{60x}} \\
  \dfrac{{1.5}}{v} = - 0.5\left( {\dfrac{{x + 120}}{{60x}}} \right) \\
  v = - \dfrac{{15}}{5}\left( {\dfrac{{60x}}{{x + 120}}} \right) \\
  v = \dfrac{{ - 180x}}{{120 + x}} \\ $
Hence, the position of the image formed after the first refraction is $ \dfrac{{ - 180x}}{{120 + x}} $ .
Now, this position acts as an object for the next reflection.
The formula to find focal length of concave lens is :
 $ f = - \dfrac{R}{2} = - \dfrac{{20}}{2} = - 10 $
We also know that the mirror formula is
 $ \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \\
  \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} \\ $
Substituting the values we will get
 $ \dfrac{1}{v} = - \dfrac{1}{{10}} + \dfrac{1}{{\dfrac{{180x}}{{120 + x}}}} \\
  \dfrac{1}{v} = - \dfrac{1}{{10}} + \dfrac{{120 + x}}{{180x}} \\
  \dfrac{1}{v} = \dfrac{{ - 18x + 120 + x}}{{180x}} \\
  \dfrac{1}{v} = \dfrac{{ - 17x + 120}}{{180x}} \\
  v = \dfrac{{180x}}{{120 - 17x}} \\ $
Now, this position acts as an object for the next refraction. So we will again use the refraction formula.
The image position should be the same as that of the position of the pin. Hence, $ v = - x $ .
 $ \dfrac{{{\mu _2}}}{v} - \dfrac{{{\mu _1}}}{u} = \dfrac{{{\mu _2} - {\mu _1}}}{R} $
Substituting the values we will get,
 $ \dfrac{1}{{ - x}} - \dfrac{{1.5}}{{180x}}\left( {120 - 17x} \right) = \dfrac{{ - 0.5}}{{60}} \\
  \dfrac{1}{{ - x}} - \dfrac{{1.5}}{{180x}}\left( {120 - 17x} \right) = \dfrac{{ - 0.5}}{{60}} \\
  \dfrac{1}{1} + \dfrac{{120 - 17x}}{{120}} = \dfrac{x}{{120}} \\
  1 = \dfrac{{ - 120 + 17x + x}}{{120}} \\
  120 = - 120 + 18x \\
  240 = 18x \\
  x = \dfrac{{240}}{{18}} \\
  x = 15 \\$
Hence, the pin should be placed on 15cm so that its image is formed at the same place
(b)
Let us consider the focal length of water as $ {f_w} $ and the focal length of the whole combination as $ F' $ . The focal length for refraction is $ {f_g} $ and reflection is $ {f_m} $ .
Hence, $ F' $ can be calculated as:
 $ \dfrac{1}{{F'}} = \dfrac{1}{{{f_w}}} + \dfrac{1}{{{f_g}}} + \dfrac{1}{{{f_m}}} + \dfrac{1}{{{f_g}}} + \dfrac{1}{{{f_w}}} \\
  \dfrac{1}{{F'}} = \dfrac{2}{{{f_w}}} + \dfrac{2}{{{f_g}}} + \dfrac{1}{{{f_m}}} \\ $
Now, to calculate the value of $ {f_w} $ , We use this formula:
 $ \dfrac{1}{{{f_w}}} = \left( {\mu _w^a - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right) \\
  \dfrac{1}{{{f_w}}} = \left( {\dfrac{4}{3} - 1} \right)\left( {\dfrac{1}{{60}}} \right) \\
  \dfrac{1}{{{f_w}}} = \dfrac{1}{{180}} \\
  {f_w} = 180cm \\ $
We also know that $ {f_g} = 60cm $ and $ {f_m} = 10cm $
Hence after substituting all the values we will get,
 $ \dfrac{1}{{F'}} = \dfrac{2}{{180}} + \dfrac{2}{{60}} + \dfrac{1}{{10}} \\
  \dfrac{1}{{F'}} = \dfrac{{2 + 6 + 18}}{{180}} \\
  \dfrac{1}{{F'}} = \dfrac{{26}}{{180}} \\
  F' = \dfrac{{90}}{{13}} \\ $
The formula to calculate u is
 $ u' = 2F' \\
  u' = \dfrac{{2 \times 90}}{{13}} = \dfrac{{180}}{{13}}cm \\ $
Now to calculate the displacement of the pin from the original position, we subtract both the values
 $ = u - u' \\
   = 15 - \dfrac{{180}}{{13}} \\
   = \dfrac{{15}}{{13}} = 1.14cm \\ $
Hence, the image should be displaced by 1.14cm so that the image of the pin again coincides with the pin.

Note:
In a concavo convex lens, the degree of curvature of the convex face is greater than that of the concave face. The lenses can be positive or negative. It depends upon the curvatures of two surfaces.