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QJEEAdv19Phfc6e: JEE Advanced 2019 Physics - Class 12 JEEAdv19Phfc6e Solution: Three glass cylinders of equal height H = 30 cm and same…

Three glass cylinders of equal height H = 30 cm and same refractive index n = 1.5 are placed on a horizontal surface as shown in figure.

JEE Advanced 2019 Paper 2 Offline Physics - Geometrical Optics Question 48 English

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 = 3 m). If H1, H2, and H3 are the apparent depths of a point X on the bottom of the three cylinders, respectively, the correct statement(s) is/are

Multiple Choice Options

  • A H2 > H1
  • B H3 > H1
  • C 0.85 cm < (H2 −- H1) < 0.9 cm
  • D H2 > H3

Step-by-Step Verified Solution

Case - I

JEE Advanced 2019 Paper 2 Offline Physics - Geometrical Optics Question 48 English Explanation 1 H = 30 cm

n = 32{3 \over 2}

H1 = Hn{H \over n}

⇒\Rightarrow 30×23{{30 \times 2} \over 3} = 20 cm

Case - II

JEE Advanced 2019 Paper 2 Offline Physics - Geometrical Optics Question 48 English Explanation 2 R = 300 cm

n2v−n1u=n2−n1R{{{n_2}} \over v} - {{{n_1}} \over u} = {{{n_2} - {n_1}} \over R}

1−H2−3−2×30=1−32−300{1 \over { - {H_2}}} - {3 \over { - 2 \times 30}} = {{1 - {3 \over 2}} \over { - 300}}

H2=60029=20.684{H_2} = {{600} \over {29}} = 20.684 cm

Case - III

JEE Advanced 2019 Paper 2 Offline Physics - Geometrical Optics Question 48 English Explanation 3 n2v−n1u=n2−n1R{{{n_2}} \over v} - {{{n_1}} \over u} = {{{n_2} - {n_1}} \over R} ;

1−H3−3−2×30=1−32300{1 \over { - {H_3}}} - {3 \over { - 2 \times 30}} = {{1 - {3 \over 2}} \over {300}}

H3=60031=19.354{H_3} = {{600} \over {31}} = 19.354 cm

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