Ray Optics

JEE Main level, single correct, medium

37 of 295

Question 37 of 295

Region II and IIII are separated by a spherical surface of radius 25cm25cm. An object is kept in region II at a distance of 40cm40cm from the surface. The distance of
the image from the surface is:

Show answer and solution

Answer: (D) 37.58cm37.58cm

μ2v−μ1u=μ2−μ1R\dfrac{{\mu }_{2}}{v}-\dfrac{{\mu }_{1}}{u}=\dfrac{{\mu }_{2}-{\mu }_{1}}{R}
1.4v−1.25−40=1.4−1.25−25\dfrac{1.4}{v}-\dfrac{1.25}{-40}=\dfrac{1.4-1.25}{-25}
1.4v=−0.1525−1.2540\dfrac{1.4}{v}=-\dfrac{0.15}{25}-\dfrac{1.25}{40}
v=−37.58cmv=-37.58cm
Hence option (4)(4).

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Lens ray diagram simulationDrag the sliders and watch the numbers in this question change.Revise Ray Optics firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Ray Optics (Mirror Formula, Reflection in Spherical Mirror), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.