JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 22)
A convex lens, of focal length 30 cm, a concave lens of focal length 120 cm, and aplane mirror are arranged as shown. For an object kept at a distance of 60 cm from the convex lens, the final image, formed by the combination, is a real image, at a distance of :
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60 cm from the convex lens
60 cm from the concave lens
70 cm from the convex lens
70 cm from the concave lens
Explanation
For convex lens,
$${1 \over {30}} = {1 \over {{V_1}}} + {1 \over {60}}$$
$$ \Rightarrow $$ $${1 \over {{V_1}}} = {1 \over {60}}$$
$$ \Rightarrow $$ $${V_1} = 60$$ cm
For concave lens,
$${1 \over { - 120}} = {1 \over {{V_2}}} - {1 \over {40}}$$
$$ \Rightarrow $$ $${1 \over {{V_2}}} = {1 \over {60}}$$
$$ \Rightarrow $$ $${V_2} = 60$$ cm
object will be at 60 cm distance from concave lens.
As mirror distance is 50 cm from concave lens. So virtual object is 10 cm behind mirror.
So, real image is 10 cm infront of mirror.
Distance of image from concave lens $$=$$ 50 $$-$$ 10 $$=$$ 40 cm. Which is same for the real object.
So similar situation will happen and final image will form at the real object it self.
$${1 \over {30}} = {1 \over {{V_1}}} + {1 \over {60}}$$
$$ \Rightarrow $$ $${1 \over {{V_1}}} = {1 \over {60}}$$
$$ \Rightarrow $$ $${V_1} = 60$$ cm
For concave lens,
$${1 \over { - 120}} = {1 \over {{V_2}}} - {1 \over {40}}$$
$$ \Rightarrow $$ $${1 \over {{V_2}}} = {1 \over {60}}$$
$$ \Rightarrow $$ $${V_2} = 60$$ cm
object will be at 60 cm distance from concave lens.
As mirror distance is 50 cm from concave lens. So virtual object is 10 cm behind mirror.
So, real image is 10 cm infront of mirror.
Distance of image from concave lens $$=$$ 50 $$-$$ 10 $$=$$ 40 cm. Which is same for the real object.
So similar situation will happen and final image will form at the real object it self.
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