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Geometrical Optics question

2025 · 2 Apr · Shift 2 · Q61
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Geometrical Optics question

2025 · 2 Apr · Shift 2 · Q61

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
A bi-convex lens has radius of curvature of both the surfaces same as 1/6 cm1 / 6 \mathrm{~cm}1/6 cm. If this lens is required to be replaced by another convex lens having different radii of curvatures on both sides (R1eqR2)\left(R_1 eq R_2\right)(R1​eqR2​), without any change in lens power then possible combination of R1R_1R1​ and R2R_2R2​ is :
  1. A
    13 cm\frac{1}{3} \mathrm{~cm}31​ cm and 17 cm\frac{1}{7} \mathrm{~cm}71​ cm
  2. B
    15 cm\frac{1}{5} \mathrm{~cm}51​ cm and 17 cm\frac{1}{7} \mathrm{~cm}71​ cm
  3. C
    13 cm\frac{1}{3} \mathrm{~cm}31​ cm and 13 cm\frac{1}{3} \mathrm{~cm}31​ cm
  4. D
    16 cm\frac{1}{6} \mathrm{~cm}61​ cm and 19 cm\frac{1}{9} \mathrm{~cm}91​ cm
View written solutionFree

Correct answer: B

  1. Use lens maker's formula

For a thin lens in air,

P=(c-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)

where PPP is the power and μ\muμ is the refractive index.

For a bi-convex lens with equal radii, the magnitudes of radii are same. Using sign convention:

  • first surface: R1=+16 cmR_1=+\frac{1}{6}\,\text{cm}R1​=+61​cm
  • second surface: R2=−16 cmR_2=-\frac{1}{6}\,\text{cm}R2​=−61​cm

So,

P=(μ−1)(11/6−1−1/6)P=(\mu-1)\left(\frac{1}{1/6}-\frac{1}{-1/6}\right)P=(μ−1)(1/61​−−1/61​) P=(μ−1)(6+6)=12(μ−1)P=(\mu-1)(6+6)=12(\mu-1)P=(μ−1)(6+6)=12(μ−1)

Thus, for the new convex lens, power must also satisfy

(μ−1)(1R1−1R2)=12(μ−1)(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=12(\mu-1)(μ−1)(R1​1​−R2​1​)=12(μ−1)

Cancelling (μ−1)(\mu-1)(μ−1),

1R1−1R2=12\frac{1}{R_1}-\frac{1}{R_2}=12R1​1​−R2​1​=12

For a convex lens, with light from left to right:

  • first surface radius is positive
  • second surface radius is negative

Hence if magnitudes are given as R1,R2R_1, R_2R1​,R2​, then condition becomes

1R1+1R2=12\frac{1}{R_1}+\frac{1}{R_2}=12R1​1​+R2​1​=12
  1. Check each option

Option A: 13\frac{1}{3}31​ cm and 17\frac{1}{7}71​ cm

1R1+1R2=3+7=10\frac{1}{R_1}+\frac{1}{R_2}=3+7=10R1​1​+R2​1​=3+7=10

Not equal to 121212. So A is incorrect.

Option B: 15\frac{1}{5}51​ cm and 17\frac{1}{7}71​ cm

1R1+1R2=5+7=12\frac{1}{R_1}+\frac{1}{R_2}=5+7=12R1​1​+R2​1​=5+7=12

This matches. So B is correct.

Option C: 13\frac{1}{3}31​ cm and 13\frac{1}{3}31​ cm

1R1+1R2=3+3=6\frac{1}{R_1}+\frac{1}{R_2}=3+3=6R1​1​+R2​1​=3+3=6

Not equal to 121212. So C is incorrect.

Option D: 16\frac{1}{6}61​ cm and 19\frac{1}{9}91​ cm

1R1+1R2=6+9=15\frac{1}{R_1}+\frac{1}{R_2}=6+9=15R1​1​+R2​1​=6+9=15

Not equal to 121212. So D is incorrect.

  1. Final answer

The possible combination is

15 cm and 17 cm\boxed{\frac{1}{5}\,\text{cm and }\frac{1}{7}\,\text{cm}}51​cm and 71​cm​

So the correct option is B.

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