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Physics · Geometrical Optics

NEET (UG) 2023 — Question 45

In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin)?

Question figure
  1. Option A:

    -40 cm

  2. Option B:

    -100 cm

    Correct
  3. Option C:

    -50 cm

  4. Option D:

    40 cm

Answer: B

Step-by-step solution

 Effective focal length ⇒feff 1feff =1f1+1f2+1f3 Also, 1f=(μ−1)(1R1−1R2)1f1=(1.6−1)(1∞−120)=−0.6201f2=(1.5−1)(120−1−20)=0.5101f3=(1.6−1)(1−20−1∞)=−0.6201feff =−0.620+0.510−0.6201feff =−0.610+0.510=−0.110=−1100∴feff =−100 cm\begin{aligned} & \text { Effective focal length } \Rightarrow f_{\text {eff }} \\ & \frac{1}{f_{\text {eff }}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3} \\ & \text { Also, } \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\ & \frac{1}{f_1}=(1.6-1)\left(\frac{1}{\infty}-\frac{1}{20}\right)=\frac{-0.6}{20} \\ & \frac{1}{f_2}=(1.5-1)\left(\frac{1}{20}-\frac{1}{-20}\right)=\frac{0.5}{10} \\ & \frac{1}{f_3}=(1.6-1)\left(\frac{1}{-20}-\frac{1}{\infty}\right)=\frac{-0.6}{20} \\ & \frac{1}{f_{\text {eff }}}=\frac{-0.6}{20}+\frac{0.5}{10}-\frac{0.6}{20} \\ & \frac{1}{f_{\text {eff }}}=\frac{-0.6}{10}+\frac{0.5}{10}=\frac{-0.1}{10}=\frac{-1}{100} \\ & \therefore f_{\text {eff }}=-100 \mathrm{~cm}\end{aligned}

Answer key and solution verified before publishing.

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Exam
NEET (UG) 2023
Paper
2023 paper
Subject
Physics
Chapter
Geometrical Optics
Topic
Lenses and Their Combinations, Silvering of Lens

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