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Geometrical Optics question

2019 · 10 Jan · Shift 2 · Q59
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Geometrical Optics question

2019 · 10 Jan · Shift 2 · Q59

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea (7.8 mm). This surface separateds two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus -
  1. A
    2 cm
  2. B
    3.1 cm
  3. C
    4.0 cm
  4. D
    1 cm
View written solutionFree

Correct answer: B

  1. Model the eye as a single spherical refracting surface

For refraction at a spherical surface, 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​​

where:

  • n1=1n_1 = 1n1​=1 (air)
  • n2=1.34n_2 = 1.34n2​=1.34 (eye medium)
  • radius of curvature R=7.8 mm=0.78 cmR = 7.8\,\text{mm} = 0.78\,\text{cm}R=7.8mm=0.78cm
  • object is at infinity, so u=−∞u = -\inftyu=−∞

Thus,

n1u=0\frac{n_1}{u} = 0un1​​=0

So the formula becomes

1.34v=1.34−10.78\frac{1.34}{v} = \frac{1.34-1}{0.78}v1.34​=0.781.34−1​
  1. Substitute and simplify
1.34v=0.340.78\frac{1.34}{v} = \frac{0.34}{0.78}v1.34​=0.780.34​

Hence,

v=1.34×0.780.34v = \frac{1.34 \times 0.78}{0.34}v=0.341.34×0.78​
  1. Calculate
v=1.04520.34≈3.074 cmv = \frac{1.0452}{0.34} \approx 3.074\,\text{cm}v=0.341.0452​≈3.074cm

Therefore,

v≈3.1 cmv \approx 3.1\,\text{cm}v≈3.1cm
  1. Match with options

The correct option is:

B: 3.1 cm3.1\,\text{cm}3.1cm

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