Question 399: Ray Optics

Question

An extended object is placed at point O, 10 cm in front of a convex lens L
1
and a concave lens L
2
is placed 10 cm behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses ae 20 cm. The refractive index of both the lenses is 1.5. The total magnification of this lens system is

Accepted Answer

Focal length of convex lens f
1
,
Therefore $${1 \over {{f_1}}} = (\mu  - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} ight] = (1.5 - 1)\left[ {{1 \over {20}} - \left( {{1 \over { - 20}}} ight)} ight]$$
$$\Rightarrow$$ f
1
= + 20 cm
Focal length of concave lens f
2
,
$$	herefore$$ $${1 \over {{f_2}}} = (\mu  - 1)\left[ {{1 \over {{R_1}}} - {1 \over {{R_2}}}} ight]$$
$${1 \over {{f_2}}} = (1.5 - 1)\left[ { - {1 \over {20}} - {1 \over {20}}} ight] = {1 \over { - 20}}$$
$$\Rightarrow$$ f
2
= $$-$$ 20 cm
For lens 1
$${1 \over v} - {1 \over u} = {1 \over f}$$
$$\Rightarrow$$ v = $$-$$ 20 cm
$${m_1} = {v \over u} = {{ - 20} \over { - 10}} = 2$$
For lens 2
u = $$-$$ 30n cm, f = $$-$$ 20 cm,
$${1 \over v} - {1 \over u} = {1 \over f}$$
v = $$-$$ 12 cm
$${m_2} = {v \over u} = {{ - 12} \over { - 30}} = {2 \over 5}$$
Net magnification,
$$m = {m_1}{m_2} = 2 	imes {2 \over 5} = {4 \over 5} = 0.8$$

Answer

0.4

Answer

0.8

Answer

1.3

Answer

1.6

Ray Optics

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