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Geometrical Optics question

2019 · Shift 2 · Q39
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Geometrical Optics question

2019 · Shift 2 · Q39

JEE AdvancedPhysicsGeometrical OpticsMultiple correct+4 / −1
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 53 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
  1. A
    H2 > H1
  2. B
    H3 > H1
  3. C
    0.85 cm < (H2 −-− H1) < 0.9 cm
  4. D
    H2 > H3
View written solutionFree

Correct answer: A, D

  1. Given data
  • Real depth of point at bottom of each cylinder: H=30 cm=0.30 mH = 30\text{ cm} = 0.30\text{ m}H=30 cm=0.30 m
  • Refractive index of glass: n=1.5n = 1.5n=1.5
  • Radius of curvature of curved top surfaces: R=3 m=300 cmR = 3\text{ m} = 300\text{ cm}R=3 m=300 cm

We need the apparent depth of the bottom point XXX when viewed from air through the top surface.


  1. Cylinder I: plane top surface

For refraction at a plane surface, apparent depth is

H1=Hn=301.5=20 cmH_1 = \frac{H}{n} = \frac{30}{1.5} = 20\text{ cm}H1​=nH​=1.530​=20 cm


  1. Cylinder II: convex top surface

This is refraction at a spherical surface from glass (n1=1.5)(n_1=1.5)(n1​=1.5) to air (n2=1)(n_2=1)(n2​=1).

Use the spherical refraction formula:

n2v−n1u=n2−n1R\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R}vn2​​−un1​​=Rn2​−n1​​

Take Cartesian sign convention with positive direction upward (towards air).

  • Object XXX is inside glass, below the top surface, so u=−30 cmu = -30\text{ cm}u=−30 cm
  • For a convex top as seen from inside glass, the center of curvature lies below the surface, so R=−300 cmR = -300\text{ cm}R=−300 cm

Substitute:

1v−1.5−30=1−1.5−300\frac{1}{v} - \frac{1.5}{-30} = \frac{1-1.5}{-300}v1​−−301.5​=−3001−1.5​

1v+0.05=−0.5−300=1600\frac{1}{v} + 0.05 = \frac{-0.5}{-300} = \frac{1}{600}v1​+0.05=−300−0.5​=6001​

1v=1600−0.05=0.0016667−0.05=−0.0483333\frac{1}{v} = \frac{1}{600} - 0.05 = 0.0016667 - 0.05 = -0.0483333v1​=6001​−0.05=0.0016667−0.05=−0.0483333

v≈−20.69 cmv \approx -20.69\text{ cm}v≈−20.69 cm

So the image is inside glass at depth 20.69 cm20.69\text{ cm}20.69 cm below the top. Hence

H2≈20.69 cmH_2 \approx 20.69\text{ cm}H2​≈20.69 cm

Therefore,

H2>H1H_2 > H_1H2​>H1​

and

H2−H1≈20.69−20=0.69 cmH_2 - H_1 \approx 20.69 - 20 = 0.69\text{ cm}H2​−H1​≈20.69−20=0.69 cm

So option C is false.


  1. Cylinder III: concave top surface

Again use

n2v−n1u=n2−n1R\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2-n_1}{R}vn2​​−un1​​=Rn2​−n1​​

Now for a concave top as seen from inside glass, the center of curvature lies above the surface, so

R=+300 cmR = +300\text{ cm}R=+300 cm

Also,

u=−30 cmu = -30\text{ cm}u=−30 cm

Substitute:

1v−1.5−30=1−1.5300\frac{1}{v} - \frac{1.5}{-30} = \frac{1-1.5}{300}v1​−−301.5​=3001−1.5​

1v+0.05=−1600\frac{1}{v} + 0.05 = -\frac{1}{600}v1​+0.05=−6001​

1v=−1600−0.05=−0.0016667−0.05=−0.0516667\frac{1}{v} = -\frac{1}{600} - 0.05 = -0.0016667 - 0.05 = -0.0516667v1​=−6001​−0.05=−0.0016667−0.05=−0.0516667

v≈−19.35 cmv \approx -19.35\text{ cm}v≈−19.35 cm

Thus,

H3≈19.35 cmH_3 \approx 19.35\text{ cm}H3​≈19.35 cm

So,

H3<H1H_3 < H_1H3​<H1​

Hence option B is false.


  1. Compare H2H_2H2​ and H3H_3H3​

We found:

H2≈20.69 cm,H3≈19.35 cmH_2 \approx 20.69\text{ cm}, \qquad H_3 \approx 19.35\text{ cm}H2​≈20.69 cm,H3​≈19.35 cm

Thus,

H2>H3H_2 > H_3H2​>H3​

So option D is true.


  1. Evaluate all options
  • A: H2>H1H_2 > H_1H2​>H1​ ✅ True
  • B: H3>H1H_3 > H_1H3​>H1​ ❌ False
  • C: 0.85 cm<(H2−H1)<0.9 cm0.85\text{ cm} < (H_2-H_1) < 0.9\text{ cm}0.85 cm<(H2​−H1​)<0.9 cm ❌ False, since H2−H1≈0.69 cmH_2-H_1 \approx 0.69\text{ cm}H2​−H1​≈0.69 cm
  • D: H2>H3H_2 > H_3H2​>H3​ ✅ True

  1. Final answer

The correct options are:

A,D\boxed{A, D}A,D​

This matches the stored correct answer.

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