Start free

Physics · Geometrical Optics

JEE Main 2026 — 8 April, Evening Shift — Question 5

A thin biconvex lens is prepared from the glass (μ=1.5)(\mu = 1.5) both curved surfaces of which have equal radii of 20cm20\mathrm{cm} each. Left side surface of the lens is silvered from outside to make it reflecting. To have the position of image and object at the same place, the object should be placed, from the lens at a distance of cm\mathrm{cm}.

  1. Option A:

    10

    Correct
  2. Option B:

    12.5

  3. Option C:

    13

  4. Option D:

    13.5

Answer: A

Step-by-step solution

Lens focal length fL=R2(μ−1)=202×0.5=20f_L = \frac{R}{2(\mu-1)} = \frac{20}{2\times0.5}=20 cm. Mirror focal length fm=−R2=−10f_m = -\frac{R}{2} = -10 cm. Combined focal length 1f=1fm−2fL=−110−220=−15\frac1f = \frac1{f_m} - \frac2{f_L} = -\frac1{10} - \frac2{20} = -\frac15 ⇒ f=−5f = -5 cm. For image at object position, object distance u=2∣f∣=10u = 2|f| = 10 cm.

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 2026
Paper
8 April, Evening Shift
Subject
Physics
Chapter
Geometrical Optics
Topic
Lenses and Their Combinations, Silvering of Lens
A thin biconvex lens is prepared from the glass (μ = 1.5) both curved… | JEE Main 2026 PYQ with Solution · DhiX AI