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Geometrical Optics question

2022 · 28 Jul · Shift 2 · Q55
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  5. /2022 · 28 Jul · Shift 2 · Q55

Geometrical Optics question

2022 · 28 Jul · Shift 2 · Q55

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
The power of a lens (biconvex) is 1.25 m−11.25 \mathrm{~m}^{-1}1.25 m−1 in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20 cm20 \mathrm{~cm}20 cm and 40 cm40 \mathrm{~cm}40 cm respectively. The refractive index of surrounding medium:
  1. A
    1.0
  2. B
    97\frac{9}{7}79​
  3. C
    32\frac{3}{2}23​
  4. D
    43\frac{4}{3}34​
View written solutionFree

Correct answer: B

  1. Use lens maker formula in a medium

For a thin lens in a surrounding medium of refractive index nmn_mnm​:

P=1f=(nlnm−1)(1R1−1R2)P=\frac{1}{f}=\left(\frac{n_l}{n_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)P=f1​=(nm​nl​​−1)(R1​1​−R2​1​)

where:

  • nl=1.5n_l=1.5nl​=1.5 is the refractive index of lens,
  • nmn_mnm​ is the refractive index of surrounding medium,
  • P=1.25 m−1P=1.25\,\text{m}^{-1}P=1.25m−1.
  1. Assign signs to radii

For a biconvex lens, taking light from left to right:

  • first surface is convex, so R1=+20 cm=+0.20 mR_1=+20\,\text{cm}=+0.20\,\text{m}R1​=+20cm=+0.20m,
  • second surface is convex toward left, so R2=−40 cm=−0.40 mR_2=-40\,\text{cm}=-0.40\,\text{m}R2​=−40cm=−0.40m.

Thus,

1R1−1R2=10.20−1−0.40=5+2.5=7.5\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{0.20}-\frac{1}{-0.40}=5+2.5=7.5R1​1​−R2​1​=0.201​−−0.401​=5+2.5=7.5
  1. Substitute into the formula
1.25=(1.5nm−1)(7.5)1.25=\left(\frac{1.5}{n_m}-1\right)(7.5)1.25=(nm​1.5​−1)(7.5)

So,

1.257.5=1.5nm−1\frac{1.25}{7.5}=\frac{1.5}{n_m}-17.51.25​=nm​1.5​−1 16=1.5nm−1\frac{1}{6}=\frac{1.5}{n_m}-161​=nm​1.5​−1 1.5nm=1+16=76\frac{1.5}{n_m}=1+\frac{1}{6}=\frac{7}{6}nm​1.5​=1+61​=67​
  1. Solve for nmn_mnm​
nm=1.5⋅67=32⋅67=97n_m=1.5\cdot \frac{6}{7}=\frac{3}{2}\cdot\frac{6}{7}=\frac{9}{7}nm​=1.5⋅76​=23​⋅76​=79​
  1. Match with the options
nm=97n_m=\frac{9}{7}nm​=79​

So the correct option is B.

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