Start free

Physics · Geometrical Optics

JEE Main 2025 — 8 April, Evening Shift — Question 65

A concave-convex lens of refractive index 1.5 and the radii of curvature of its surfaces are 30 cm and 20 cm , respectively. The concave surface is upwards and is filled with a liquid of refractive index 1.3. The focal length of the liquid-glass combination will be

Question figure
  1. Option A:

    60011 cm\frac{600}{11} \mathrm{~cm}

    Correct
  2. Option B:

    50011 cm\frac{500}{11} \mathrm{~cm}

  3. Option C:

    70011 cm\frac{700}{11} \mathrm{~cm}

  4. Option D:

    80011 cm\frac{800}{11} \mathrm{~cm}

Answer: A

Step-by-step solution

1F=1F1+1F21F1=1.3−1(−(−130))1F=1120+1100=10+121200=0.330=1100\begin{array}{ll} \frac{1}{F}=\frac{1}{F_{1}}+\frac{1}{F_{2}} \\ & \frac{1}{F_{1}}=1.3-1\left(-\left(\frac{-1}{30}\right)\right) \frac{1}{F}=\frac{1}{120}+\frac{1}{100}=\frac{10+12}{1200} \\ & =\frac{0.3}{30}=\frac{1}{100} \end{array} 1F=221200\frac{1}{F}=\frac{22}{1200} 1F2=(1.5−1)(−130−1−20)\frac{1}{F_{2}}=(1.5-1)\left(\frac{-1}{30}-\frac{1}{-20}\right) F=60011F=\frac{600}{11} =0.5(360−260)=0.5\left(\frac{3}{60}-\frac{2}{60}\right) =1120=\frac{1}{120}

Answer key and solution verified before publishing.

Practise Geometrical Optics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2025
Paper
8 April, Evening Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Lenses and Their Combinations, Silvering of Lens

More Geometrical Optics questions from this paper

A concave-convex lens of refractive index 1.5 and the radii of… | JEE Main 2025 PYQ with Solution · DhiX AI