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Physics · Geometrical Optics

JEE Advanced 2021 — Paper 1 — Question 2

An extended object is placed at point O,10 cm\mathrm{O}, 10 \mathrm{~cm} in front of a convex lens L1\mathrm{L}_{1} and a concave lens L2\mathrm{L}_{2}

is placed 10 cm behind it, as shown in the figure. The radii of curvature of all the curved surfaces in both the lenses are 20 cm .

The refractive index of both the lenses is 1.5 . The total magnification of this lens system is

Question figure
  1. Option A:

    0.4

  2. Option B:

    0.8

    Correct
  3. Option C:

    1.3

  4. Option D:

    1.6

Answer: B

Step-by-step solution

1f1=(1.5−1)(120+120)=120\frac{1}{\mathrm{f}_{1}}=(1.5-1)\left(\frac{1}{20}+\frac{1}{20}\right)=\frac{1}{20} 1f2=(1.5−1)(−120−120)=−120\frac{1}{\mathrm{f}_{2}}=(1.5-1)\left(-\frac{1}{20}-\frac{1}{20}\right)=-\frac{1}{20}

So, 1v−1−10=120\frac{1}{\mathrm{v}}-\frac{1}{-10}=\frac{1}{20}

So, v = - 20 cm and 1v′−1−30=1−20\frac{1}{v^{\prime}}-\frac{1}{-30}=\frac{1}{-20}

So, v′=−12 cm\mathrm{v}^{\prime}=-12 \mathrm{~cm}

So total magnification =(−20−10)(−12−30)=0.8=\left(\frac{-20}{-10}\right)\left(\frac{-12}{-30}\right)=0.8

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2021
Paper
Paper 1
Subject
Physics
Chapter
Geometrical Optics
Topic
Miscellaneous Problems for different combinations
An extended object is placed at point O , 10 cm in front of a convex… | JEE Advanced 2021 PYQ with Solution · DhiX AI