JEE MAIN - Physics (2019 - 11th January Morning Slot - No. 22)
An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5 m/s, the speed and direction of the image will be :
1.16 $$ \times $$ 10–3 m/s towards the lens
2.26 $$ \times $$ 10–3 m/s away from the lens
3.22 × 10–3 m/s towards the lens
0.92 $$ \times $$ 10$$-$$3 m/s away from the lens
Explanation
From lens equation
$${1 \over v} - {1 \over u} = {1 \over f}$$
$${1 \over v} - {1 \over {\left( { - 20} \right)}} = {1 \over {\left( {.3} \right)}} = {{10} \over 3}$$
$${1 \over v} = {{10} \over 3} - {1 \over {20}}$$
$${1 \over v} = {{197} \over {60}};v = {{60} \over {197}}$$
m = $$\left( {{v \over u}} \right)$$ = $${{\left( {{{60} \over {197}}} \right)} \over {20}}$$
velocity of image wrt. to lens is given by
vI/L = m2vO/L
direction of velocity of image is same as that of object
vO/L = 5 m/s
$${1 \over v} - {1 \over u} = {1 \over f}$$
$${1 \over v} - {1 \over {\left( { - 20} \right)}} = {1 \over {\left( {.3} \right)}} = {{10} \over 3}$$
$${1 \over v} = {{10} \over 3} - {1 \over {20}}$$
$${1 \over v} = {{197} \over {60}};v = {{60} \over {197}}$$
m = $$\left( {{v \over u}} \right)$$ = $${{\left( {{{60} \over {197}}} \right)} \over {20}}$$
velocity of image wrt. to lens is given by
vI/L = m2vO/L
direction of velocity of image is same as that of object
vO/L = 5 m/s
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