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JEE Main 2025 (7 Apr, Shift 1) · Refraction, Glass Slab & Apparent Depth

Refraction, Glass Slab & Apparent Depth question from JEE Main 2025 (7 Apr, Shift 1)

Q15JEE Main 2025 · 7 APR · SHIFT 1Medium

Refraction, Glass Slab & Apparent Depth

A container contains a liquid with refractive index of 1.2 up to a height of 60 cm and another liquid having refractive index 1.6 is added to height H above first liquid. If viewed from above, the apparent shift in the position of bottom of container is 40 cm. The value of H is ________ cm. (Consider liquids are immisible)

Numerical answer type. Work it out, then open the answer.

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Answer: 80

  1. Shift of bottom =∑ti(1−1μi)= \sum t_i\left(1 - \frac{1}{\mu_i}\right).
  2. Lower liquid: 60(1−11.2)=60×16=10 cm60\left(1 - \frac{1}{1.2}\right) = 60 \times \frac{1}{6} = 10\text{ cm}
  3. Upper liquid: H(1−11.6)=3H8H\left(1 - \frac{1}{1.6}\right) = \frac{3H}{8}
  4. 10+3H8=40⇒H=8010 + \frac{3H}{8} = 40 \Rightarrow H = 80
  5. So H=80 cmH = 80\text{ cm}.
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