JEE MAIN - Physics (2019 - 9th January Morning Slot - No. 26)
A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is :
1.1 cm away from the lens
0
0.55 cm towards the lens
0.55 cm away from the lens
Explanation
_9th_January_Morning_Slot_en_26_1.png)
Image is formed on the screen.
So, v = 10 cm
and $$\mu $$ = $$-$$ 10 cm
Using formula,
$${1 \over v} - {1 \over u} = {1 \over f}$$
$$ \Rightarrow $$ $${1 \over {10}} - {1 \over {\left( { - 10} \right)}}$$ = $${1 \over f}$$
$$ \Rightarrow $$ f = 5 cm
Now a glass block is placed like this,
_9th_January_Morning_Slot_en_26_2.png)
Because of this glass block source will move t$$\left( {1 - {1 \over \mu }} \right)$$ in the direction of incident ray.
$$ \therefore $$ S' = t$$\left( {1 - {1 \over \mu }} \right)$$
= 1.5$$\left( {1 - {2 \over 3}} \right)$$
= 0.5
$$ \therefore $$ now distance of source from the lens = 10 $$-$$ 0.5 = 9.5 cm
$$ \therefore $$ $$\mu $$ = $$-$$ 9.5 cm
$$ \therefore $$ $${1 \over v} - {1 \over {\left( { - 9.5} \right)}}$$ = $${1 \over 5}$$
$$ \Rightarrow $$ $${1 \over v}$$ = $${1 \over 5} - {2 \over {19}}$$
$$ \Rightarrow $$ $${1 \over v}$$ = $${9 \over {95}}$$
$$ \Rightarrow $$ v = 10.55 cm
So, to get sharp image screen should shift away (10.55 $$-$$ 10) = 0.55 cm from the lens.
Comments (0)
