QJEEM19Phbe4f: JEE Main 2019 Physics - Class 12 JEEM19Phbe4f Solution: A convex lens is put 10 cm from a light source and it makes a…

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,



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.