Question 349: Ray Optics

Question

An equiconvex lens is cut into two halves along (i)  XOX'  and  (ii)  YOY' as shown in the figure. Let $$f,f',$$ $$f''$$ be the focal lengths of the complete lens, of each half in case (i), and of each half in case (ii), respectively.
Choose the correct statement from the following

Accepted Answer

We know from Lens maker's formula
$${1 \over f} = \left( {\mu  - 1} ight)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} ight)$$
Initially, Here R
1
= R and R
2
= Ã¢ÂÂR by convection.
$$ 	herefore $$ $${1 \over f} = \left( {\mu  - 1} ight){2 \over R}$$
$$ \Rightarrow $$ $${1 \over {2f}} = \left( {\mu  - 1} ight){1 \over R}$$
If we cut the lens along XOX' then the two halves of the lens will be having the same radii of curvature and so, focal length f' = f .
But when we cut it along YOY' then, we will have
R
1
= R but R
2
= $$\infty $$
$$ 	herefore $$ $${1 \over {f''}} = \left( {\mu  - 1} ight)\left( {{1 \over R} - {1 \over \infty }} ight)$$ = $$\left( {\mu  - 1} ight){1 \over R}$$ = $${1 \over {2f}}$$
$$ \Rightarrow $$ f'' = 2f

Answer

$$f' = f,f'' = 2f$$

Answer

$$f' = 2f,f'' = f$$

Answer

$$f' = f,f'' = f$$

Answer

$$f' = 2f,f'' = 2f$$

Ray Optics

An equiconvex lens is cut into two halves along (i) XOX' and (ii) YOY' as shown in the figure. Let $f,f', f''$ be the focal lengths of the complete lens, of each half in case (i), and of each half in case (ii), respectively. Choose the correct statement from the following

Choose the correct answer: