JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Geometrical Optics

An object and a concave mirror of focal length

f = 10 c m

both move along the principal axis of the mirror with constant speeds. The object moves with speed

V_{0} = 15 c m s^{- 1}

towards the mirror with respect to a laboratory frame. The distance between the object and the mirror at a given moment is denoted by

u

. When

u = 30 c m

, the speed of the mirror

V_{m}

is such that the image is instantaneously at rest with respect to the laboratory frame, and the object forms a real image. The magnitude of

V_{m}

is ____

c m s^{- 1}

Answer: 3

Solution

Using the mirror formula:

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

With

u = - 30

cm and

f = - 10

cm:

v = - 15

cm

The velocity relation for a moving mirror:

v_{I} - v_{m} = - (\frac{v}{u})^{2} (v_{O} - v_{m})

Given

v_{I} = 0

(image at rest):

- v_{m} = - (\frac{- 15}{- 30})^{2} (v_{O} - v_{m}) = - \frac{1}{4} (15 - v_{m})

Solving:

- 4 v_{m} = - 15 + v_{m}

, giving

v_{m} = 3

cm/s

Therefore,

∣ V_{m} ∣ = 3

cm/s

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Integer answer