Question 55: Ray Optics

Question

Two identical symmetric double convex lenses of focal length f are cut into two equal parts L
1
, L
2
by AB plane and L
3
, L
4
by XY plane as shown in figure respectively. The ratio of focal lengths of lenses L
1
and L
3
is

Accepted Answer

Here,
we will use lens maker formula,
$${1 \over f} = \left( {\mu  - 1} ight)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} ight)$$
So, the focal length of L$_1$ remains the same as the original lens because the curvature of the surfaces does not change
So, $${\left( f ight)_{{L_1}}} = f$$ .... (1)
Now, for L$_3$, L$_3$ is plano-convex lens
$${R_1} = R,\,{R_2} = \infty $$
So, $${1 \over {{{\left( f ight)}_{{L_3}}}}} = \left( {\mu  - 1} ight)\left( {{1 \over R} - {1 \over \infty }} ight)$$
$$ \Rightarrow {1 \over {{{\left( f ight)}_{{L_3}}}}} = \left( {\mu  - 1} ight){1 \over R}$$ ..... (2)
For whole double convex lens,
$${1 \over f} = \left( {\mu  - 1} ight)\left( {{1 \over R} - {1 \over { - R}}} ight)$$ (As $${R_1} = R$$, $${R_2} =  - R$$)
$$ \Rightarrow {1 \over f} = \left( {\mu  - 1} ight)\left( {{1 \over R} + {1 \over R}} ight)$$
$$ \Rightarrow {1 \over f} = \left( {\mu  - 1} ight)\left( {{2 \over R}} ight)$$ .... (3)
From (2) and (3),
$${1 \over {{{\left( f ight)}_{{L_3}}}}} = {1 \over {2f}} \Rightarrow {\left( f ight)_{{L_3}}} = 2f$$
Hence, $${{{{\left( f ight)}_{{L_1}}}} \over {{{\left( f ight)}_{{L_3}}}}} = {f \over {2f}} \quad	ext{....(from (1)})\quad= 1:2$$

Answer

1 : 2

Answer

1 : 1

Answer

2 : 1

Answer

1 : 4

Ray Optics

Two identical symmetric double convex lenses of focal length f are cut into two equal parts L 1 , L 2 by AB plane and L 3 , L 4 by XY plane as shown in figure respectively. The ratio of focal lengths of lenses L 1 and L 3 is

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