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Physics · Geometrical Optics

JEE Advanced 2019 — Paper 2 — Question 5

Three glass cylinders of equal height H=30 cm\mathrm{H}=30 \mathrm{~cm} and same refractive index n=1.5\mathrm{n}=1.5 are placed on a horizontal surface as shown in figure. Cylinder I has a flat top, cylinder II has a convex top and cylinder III has a concave top. The radii of curvature of the two curved tops are same (R=3m)(R=3 m). If H1,H2H_{1}, H_{2} and H3H_{3} are the apparent depths of a point X on the bottom of the three cylinders, respectively, the correct statement(s) is/are:

Question figure
  1. Option A:

    H2>H1\mathrm{H}_{2}>\mathrm{H}_{1}

    Correct
  2. Option B:

    H2>H3\mathrm{H}_{2}>\mathrm{H}_{3}

    Correct
  3. Option C:

    H3>H1\mathrm{H}_{3}>\mathrm{H}_{1}

  4. Option D:

    0.8 cm<(H2−H1)<0.9 cm0.8 \mathrm{~cm}<\left(\mathrm{H}_{2}-\mathrm{H}_{1}\right)<0.9 \mathrm{~cm}

Answer: A, B

Step-by-step solution

We apply, μ2v−μ1u=μ2−μ1R\frac{\mu_{2}}{\mathrm{v}}-\frac{\mu_{1}}{\mathrm{u}}=\frac{\mu_{2}-\mu_{1}}{\mathrm{R}}.

1H1−1.5−30=1−1.5∞…(i)\frac{1}{\mathrm{H}_{1}}-\frac{1.5}{-30}=\frac{1-1.5}{\infty} …(i)

⇒H1=20 cm\Rightarrow \mathrm{H}_{1}=20 \mathrm{~cm}

1H2−1.5−30=−0.5−300…(ii)\frac{1}{\mathrm{H}_{2}}-\frac{1.5}{-30}=\frac{-0.5}{-300} …(ii)

⇒H2=30014.5 cm\Rightarrow \mathrm{H}_{2}=\frac{300}{14.5} \mathrm{~cm}

1−H3−1.5−30=−0.5300…(iii)\frac{1}{-\mathrm{H}_{3}}-\frac{1.5}{-30}=\frac{-0.5}{300} …(iii)

⇒H3=30015.5 cm\Rightarrow \mathrm{H}_{3}=\frac{300}{15.5} \mathrm{~cm}

∴\therefore (A) and (B) are correct.

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2019
Paper
Paper 2
Subject
Physics
Chapter
Geometrical Optics
Topic
Apparent Depth, Glass Slab, Prism and Dispersion

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