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Geometrical Optics question

2025 · 29 Jan · Shift 2 · Q66
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Geometrical Optics question

2025 · 29 Jan · Shift 2 · Q66

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
Two identical symmetric double convex lenses of focal length f are cut into two equal parts L1, L2 by AB plane and L3, L4 by XY plane as shown in figure respectively. The ratio of focal lengths of lenses L1 and L3 is JEE Main 2025 (Online) 29th January Evening Shift Physics - Geometrical Optics Question 39 English
  1. A
    1 : 2
  2. B
    1 : 1
  3. C
    2 : 1
  4. D
    1 : 4
View written solutionFree

Correct answer: A

  1. Use lens maker's formula for the original double convex lens

For a thin symmetric double convex lens in air,

1f=(μ−1)(1R1−1R2)\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)f1​=(μ−1)(R1​1​−R2​1​)

For a symmetric biconvex lens, R1=+R,R2=−RR_1=+R,\qquad R_2=-RR1​=+R,R2​=−R So,

1f=(μ−1)(1R−1−R)=(μ−1)(2R)\frac{1}{f}=(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right)= (\mu-1)\left(\frac{2}{R}\right)f1​=(μ−1)(R1​−−R1​)=(μ−1)(R2​)

Hence,

1f=2(μ−1)R\frac{1}{f}=\frac{2(\mu-1)}{R}f1​=R2(μ−1)​
  1. Lens L1L_1L1​: cut by plane ABABAB

The plane ABABAB is along the principal axis, so the lens is cut into two equal halves without changing the curvatures of the refracting surfaces.

Thus each half still has the same two spherical surfaces as the original lens: R1=+R,R2=−RR_1=+R,\qquad R_2=-RR1​=+R,R2​=−R Therefore,

1f1=(μ−1)(1R−1−R)=2(μ−1)R\frac{1}{f_1}=(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right)=\frac{2(\mu-1)}{R}f1​1​=(μ−1)(R1​−−R1​)=R2(μ−1)​

So,

f1=ff_1=ff1​=f
  1. Lens L3L_3L3​: cut by plane XYXYXY

The plane XYXYXY is perpendicular to the principal axis, so the lens is cut into two parts such that one part becomes a plano-convex lens.

For this plano-convex lens,

  • one surface is spherical: R1=+RR_1=+RR1​=+R
  • the other surface is plane: R2=∞R_2=\inftyR2​=∞

Using lens maker's formula,

1f3=(μ−1)(1R−1∞)\frac{1}{f_3}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)f3​1​=(μ−1)(R1​−∞1​)

Since 1∞=0\frac{1}{\infty}=0∞1​=0,

1f3=μ−1R\frac{1}{f_3}=\frac{\mu-1}{R}f3​1​=Rμ−1​

But from the original lens,

1f=2(μ−1)R\frac{1}{f}=\frac{2(\mu-1)}{R}f1​=R2(μ−1)​

Therefore,

1f3=12f\frac{1}{f_3}=\frac{1}{2f}f3​1​=2f1​

So,

f3=2ff_3=2ff3​=2f
  1. Find the ratio

We have,

f1=f,f3=2ff_1=f,\qquad f_3=2ff1​=f,f3​=2f

Hence,

f1:f3=f:2f=1:2f_1:f_3 = f:2f = 1:2f1​:f3​=f:2f=1:2
  1. Check options
  • A: 1:21:21:2 ✅
  • B: 1:11:11:1 ❌
  • C: 2:12:12:1 ❌
  • D: 1:41:41:4 ❌

So the correct option is A.

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