
Two thin similar convex glass pieces are joined together front to front with its rear portion silvered such that a sharp image is formed $20\,cm$ from the mirror. When the air between the glass pieces is replaced by water $\left( {{\mu _w} = \dfrac{4}{3}} \right)$ , then the image formed from the mirror is at a distance of
A) $8\,cm$
B) $10\,cm$
C) $6\,cm$
D) $12\,cm$
Answer
243.3k+ views
Hint: Use the formula (2) to find the radius of curvature. Substitute the radius of curvature and the refractive index in the formula (1) to find the value of focal length. Then substitute this in the other formula, to find the value of distance of image from mirror.
Formula used:
(1) The relation of the concave lenses is
$\dfrac{1}{F} = \dfrac{{4\left( {\mu - 1} \right)}}{R} + \dfrac{2}{R}$
Where $F$ is the focal length of the mirror, $\mu $ is the refractive index of the mirror, $R$ is the radius of the curvature of the mirror.
(2) For the convex mirror,
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{2}{R}$
Where $v$ is the distance of the image, $u$n is the distance of the object and $R$ is the radius of curvature.
Complete step by step solution:
It is given that the
Distance of the image formed from the object, $R = 20\,cm$
Before inserting water, it behaves as a convex mirror. Hence using formula (2)
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{2}{R}$
Substituting the values.
\[\Rightarrow \dfrac{1}{{10}} + \dfrac{1}{{10}} = \dfrac{2}{R}\]
$\Rightarrow R = 20\,cm$
After converting into concave mirror by introducing water,
$\Rightarrow \dfrac{1}{F} = \dfrac{{4\left( {\mu - 1} \right)}}{R} + \dfrac{2}{R}$
Substituting the known values in the above equation, we get
$\Rightarrow \dfrac{1}{F} = \dfrac{{4\left( {\dfrac{4}{3} - 1} \right)}}{{20}} + \dfrac{2}{{20}}$
By simplifying the above equation, we get
$\Rightarrow \dfrac{1}{F} = \dfrac{1}{6}$
We know that $\dfrac{1}{F} = \dfrac{1}{u} + \dfrac{1}{v}$ , hence substituting the values in it, we get
$\Rightarrow \dfrac{2}{u} = \dfrac{1}{F} = \dfrac{1}{6}$
$\Rightarrow u = 12\,cm$
Thus the option (D) is correct.
Note: When the air is introduced between two glass pieces, the glass pieces convert into and act as a convex mirror. But if some amount of water is placed between the two glass pieces instead of air, then it might behave as a concave mirror.
Formula used:
(1) The relation of the concave lenses is
$\dfrac{1}{F} = \dfrac{{4\left( {\mu - 1} \right)}}{R} + \dfrac{2}{R}$
Where $F$ is the focal length of the mirror, $\mu $ is the refractive index of the mirror, $R$ is the radius of the curvature of the mirror.
(2) For the convex mirror,
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{2}{R}$
Where $v$ is the distance of the image, $u$n is the distance of the object and $R$ is the radius of curvature.
Complete step by step solution:
It is given that the
Distance of the image formed from the object, $R = 20\,cm$
Before inserting water, it behaves as a convex mirror. Hence using formula (2)
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{2}{R}$
Substituting the values.
\[\Rightarrow \dfrac{1}{{10}} + \dfrac{1}{{10}} = \dfrac{2}{R}\]
$\Rightarrow R = 20\,cm$
After converting into concave mirror by introducing water,
$\Rightarrow \dfrac{1}{F} = \dfrac{{4\left( {\mu - 1} \right)}}{R} + \dfrac{2}{R}$
Substituting the known values in the above equation, we get
$\Rightarrow \dfrac{1}{F} = \dfrac{{4\left( {\dfrac{4}{3} - 1} \right)}}{{20}} + \dfrac{2}{{20}}$
By simplifying the above equation, we get
$\Rightarrow \dfrac{1}{F} = \dfrac{1}{6}$
We know that $\dfrac{1}{F} = \dfrac{1}{u} + \dfrac{1}{v}$ , hence substituting the values in it, we get
$\Rightarrow \dfrac{2}{u} = \dfrac{1}{F} = \dfrac{1}{6}$
$\Rightarrow u = 12\,cm$
Thus the option (D) is correct.
Note: When the air is introduced between two glass pieces, the glass pieces convert into and act as a convex mirror. But if some amount of water is placed between the two glass pieces instead of air, then it might behave as a concave mirror.
Recently Updated Pages
JEE Main 2026 Session 2 City Intimation Slip & Exam Date: Expected Date, Download Link

JEE Main 2026 Session 2 Application Form: Reopened Registration, Dates & Fees

JEE Main 2026 Session 2 Registration (Reopened): Last Date, Fees, Link & Process

WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

JEE Main 2025-26 Mock Tests: Free Practice Papers & Solutions

JEE Main 2025-26 Experimental Skills Mock Test – Free Practice

Trending doubts
JEE Main 2026: Session 1 Results Out and Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Understanding the Angle of Deviation in a Prism

Understanding Differential Equations: A Complete Guide

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Understanding the Block and Tackle System

