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Geometrical Optics question

2024 · 31 Jan · Shift 2 · Q86
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Geometrical Optics question

2024 · 31 Jan · Shift 2 · Q86

JEE MainPhysicsGeometrical OpticsNumerical+4 / −1
Light from a point source in air falls on a convex curved surface of radius 20 cm20 \mathrm{~cm}20 cm and refractive index 1.5. If the source is located at 100 cm100 \mathrm{~cm}100 cm from the convex surface, the image will be formed at ‾\underline{\hspace{2cm}}​cm\mathrm{cm}cm from the object.
Numerical answer
View written solutionFree

Correct answer: 200

  1. Given data
  • Refractive index of air: n1=1n_1 = 1n1​=1
  • Refractive index of curved medium: n2=1.5n_2 = 1.5n2​=1.5
  • Radius of convex spherical surface: R=20 cmR = 20\,\text{cm}R=20cm
  • Object distance from surface: u=−100 cmu = -100\,\text{cm}u=−100cm

Here, using the Cartesian sign convention, light travels from left to right, so the object is on the left of the surface and hence u<0u<0u<0.

For a convex surface as seen from the object side, the center of curvature lies on the refracted side, so

R=+20 cmR = +20\,\text{cm}R=+20cm

  1. Use refraction at a spherical surface formula

The formula is

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

Substitute the values:

1.5v−1−100=1.5−120\frac{1.5}{v} - \frac{1}{-100} = \frac{1.5-1}{20}v1.5​−−1001​=201.5−1​ 1.5v+1100=0.520=140\frac{1.5}{v} + \frac{1}{100} = \frac{0.5}{20} = \frac{1}{40}v1.5​+1001​=200.5​=401​

So,

1.5v=140−1100\frac{1.5}{v} = \frac{1}{40} - \frac{1}{100}v1.5​=401​−1001​

Take LCM:

140−1100=5−2200=3200\frac{1}{40} - \frac{1}{100} = \frac{5-2}{200} = \frac{3}{200}401​−1001​=2005−2​=2003​

Hence,

1.5v=3200\frac{1.5}{v} = \frac{3}{200}v1.5​=2003​

Since 1.5=321.5 = \frac{3}{2}1.5=23​,

3/2v=3200\frac{3/2}{v} = \frac{3}{200}v3/2​=2003​ 12v=1200\frac{1}{2v} = \frac{1}{200}2v1​=2001​ v=100 cmv = 100\,\text{cm}v=100cm

So the image is formed 100 cm100\,\text{cm}100cm to the right of the surface.

  1. Distance of image from the object

The object is 100 cm100\,\text{cm}100cm to the left of the surface, and the image is 100 cm100\,\text{cm}100cm to the right of the surface.

Therefore, distance between object and image is

100+100=200 cm100 + 100 = 200\,\text{cm}100+100=200cm
  1. Final answer

200\boxed{200}200​

  1. Comparison with stored answer

Stored correct answer = 200200200

Our derived answer also = 200200200, so they agree.

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