Physics Circle home Student Login

Home›Chapters›Ray Optics PYQs›Refraction, Glass Slab & Apparent Depth›Question 30

MHT CET 2024 (16 May, Shift 2) · Refraction, Glass Slab & Apparent Depth

Refraction, Glass Slab & Apparent Depth question from MHT CET 2024 (16 May, Shift 2)

Q30MHT CET 2024 · 16 MAY · SHIFT 2Difficult

Refraction, Glass Slab & Apparent Depth

A plane mirror is placed at the bottom of a tank containing a liquid of refractive index ' μ\mu ', ' pp ' is a small object at a height ' hh ' above the mirror. An observer ' OO ' vertically above ' pp ' outside the liquid sees ' pp ' and the image in a mirror. The apparent distance between these two will be

Figure

Show answer and solution

Answer: (b)

  1. Let the liquid surface be at depth dd above pp. The mirror image of pp is hh below the mirror, i.e. at depth d+2hd + 2h.
  2. Apparent depths: pp at dμ\frac{d}{\mu}, image at d+2hμ\frac{d + 2h}{\mu}.
  3. Apparent separation =d+2hμ−dμ=2hμ= \frac{d + 2h}{\mu} - \frac{d}{\mu} = \frac{2h}{\mu}
  4. So the answer is (b) 2hμ\frac{2h}{\mu}.
Report a mistake

More questions on Refraction, Glass Slab & Apparent Depth