Question 109: Ray Optics

Question

Two identical symmetric double convex lenses of focal length f are cut into two equal parts L1, L2 by AB plane and L3, L4 by XY plane as shown in figure respectively. The ratio of focal lengths of lenses L1 and L3 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 L1, L2 by AB plane and L3, L4 by XY plane as shown in figure respectively. The ratio of focal lengths of lenses L1 and L3 is

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