Question

18. Two converging glass lenses 'A' and 'B'have
focal lengths in the ratio 2:1. The radius of
curvature of first surface of lens ‘A’is 1/4th of
the second surface where as the radius of
curvature of first surface of lens 'B' is twice
that of second surface. Then the ratio between
the radii of the first surfaces of A and B is
1)5:3 2)3:5 3) 1:2 4 )5 6​

Answer 1

Answer:

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Answer 2

Given :

The ratio of focal lengths of two converging glass  lenses 'A' and 'B' is :

= 2:1

The radius of curvature of first surface of lens 'A' is :

= $\frac{1}{4}$ th of radius of curvature of 2nd surface

The radius of curvature of 1st surface of lens 'B' is :

= Twice the radius of curvature of 2nd surface

To Find :

Ratio between the radii of the 1st surfaces of A and B is = ?

Solution :

By Lens Maker's formula , for a lens of material of refractive index 'n', we have :

$\frac{1}{f}= (n-1) ( \frac{1}{R_1} - \frac{1}{R_2}} )$

Here, $R_1$ and $R_2$ are the radii of curvatures of 1st and 2nd surfaces respectively ,

And f = focal length of the lens

So, for the lens 'A' we have :

$\frac{1}{f_A} = (n-1) (\frac{1}{(R_1)_A}-\frac{1}{(R_2)_A})$

∴$(R_1)_A > 0$ and $(R_2)A < 0$ and $(R_1)_A= \frac{(R_2)_A}{4}$

So, $\frac{1}{f_A} = (n-1) (\frac{1}{(R_1)_A}+\frac{1}{4(R_1)_A})$

          $=(n-1)(\frac{5}{4(R_1)_A})$

Similarly for lens 'B' we have :

$\frac{1}{f_b} = (n-1) (\frac{1}{(R_1)_B}-\frac{1}{(R_2)_B})$

   $= (n-1) (\frac{1}{(R_1)_B}+\frac{1}{(\frac{(R_1)_B}{2})})$

  $=(n-1)(\frac{3}{(R_1)_B})$

So,

$\frac{f_A }{f_B} = \frac{(\frac{1}{f_B})}{(\frac{1}{f_A})}$

$\frac{2}{1}= \frac{(n-1)\frac{3}{(R_1)_B}}{(n-1)\frac{5}{4(R_1)_A}}$

$\frac{2}{1}= \frac{12}{5} \times \frac{(R_1)_A}{(R_1)_B}$

∴$\frac{(R_1)_A}{(R_1)_B}= \frac{10}{12}= 5:6$

So, the correct answer for the ratio of the radii of the 1st surfaces of A and B is option (4) . 5:6