QJEEAdv16Ph65d5: JEE Advanced 2016 Physics - Class 12 JEEAdv16Ph65d5 Solution: A small object is placed 50 cm to the left of a thin convex…

For convex lens, u = $-$ 50 cm, f = 30 cm

$\therefore$ ${1 \over v} = {1 \over f} + {1 \over u} = {1 \over {30}} - {1 \over {50}} = {{5 - 3} \over {150}} = {2 \over {150}}$ or v = 75 cm

Image formed by convex lens acts as a virtual object for mirror.

If we consider that the mirror is not tilted, then for mirror,

${u_1} = 75 - 50 = 25$ cm, ${f_1} = {R \over 2} = {{100\,cm} \over 2} = 50$ cm

$\therefore$ ${1 \over {{v_1}}} = {1 \over {{f_1}}} - {1 \over {{u_1}}} = {1 \over {50}} - {1 \over {25}} = {{ - 1} \over {50}}$ or ${v_1} = 50$ cm

So, the co-ordinates of final image formed (I₂) if the mirror is not tilted are (0 cm, 0 cm)

If mirror is not tilted then this ray starts from the object, refracts through the lens without deviation, incident normally on the mirror and finally comes to the image I₂ after reflection from the mirror. If the mirror is tilted by an angle $\theta$ then this ray is incident on the mirror at an angle of reflection $\theta$ and finally forms a new image at I'₂ (see figure). Thus, the axis on which image lies makes an angle 2 $\theta$ = 60 $^\circ$ with the principal axis of the lens. Also, the image distance will remain same upto the first order of approximation (image distance will remain exactly same in case of plane mirror). Thus, (x, y) coordinates of the image are

$x = 50 - 50\cos 60^\circ = 25$ cm,

$y = 50\sin 60^\circ = 25\sqrt 3$ cm.