
The diagram shows a mirror system. If after two reflections the image and the object O coincide with each other, then the value of the radius of curvature of the concave mirror is: (take $a = 3m$)

(A) $4m$
(B) $6m$
(C) $8m$
(D) $10m$
Answer
144.6k+ views
Hint The radius of the curvature of a mirror is twice of its focal length. Focal length can easily be calculated if the image and object position is given. For a plane mirror, the object distance is equal to image distance.
Step by step solution
Two conditions are given for this problem. First, the object is kept at O, i.e. a distance $4a$ from the concave mirror. Second, the final image after reflection from the plane mirror forms at O i.e. at a distance of $a$ behind the plane mirror.
Since object distance and image distance are always equal for a plane mirror, we can say that the object for the plane mirror is placed at position $P$
Using the given condition that reflection takes place twice, we can conclude that first, it reflects from the concave mirror and then from the plane mirror to make the final image at O. This implies that the object for the plane mirror is actually the image from the concave mirror.
So, we have the object distance and image distance for the concave mirror.
$u = - 4a$
And, $v = - 2a$.
Where, $u$ object distance
$v$ is the image distance.
The negative sign represents that these coordinates are on the left side of the pole of the mirror.
Using mirror formula,
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
Putting the values, we get
$\dfrac{1}{{ - 2a}} + \dfrac{1}{{ - 4a}} = \dfrac{1}{f}$
$ \Rightarrow \dfrac{{ - 2 - 1}}{{4a}} = \dfrac{1}{f}$
Solving this
$\dfrac{1}{f} = \dfrac{{ - 3}}{{4a}}$
We know that, radius of curvature of a mirror is given by twice its focal length.
$R = 2f$
$ \Rightarrow R = 2\left( {\dfrac{{ - 4a}}{3}} \right)$
Putting $a = 3m$, we get
$R = 8m$.
Hence, Option (C) is correct.
Note The basic concept is to remember the mirror formulas. You should be careful if you are taking the numeric value of the image and object distance because it can easily lead you to error. A better way is to solve it first and then put the numeric value.
Step by step solution
Two conditions are given for this problem. First, the object is kept at O, i.e. a distance $4a$ from the concave mirror. Second, the final image after reflection from the plane mirror forms at O i.e. at a distance of $a$ behind the plane mirror.
Since object distance and image distance are always equal for a plane mirror, we can say that the object for the plane mirror is placed at position $P$
Using the given condition that reflection takes place twice, we can conclude that first, it reflects from the concave mirror and then from the plane mirror to make the final image at O. This implies that the object for the plane mirror is actually the image from the concave mirror.
So, we have the object distance and image distance for the concave mirror.
$u = - 4a$
And, $v = - 2a$.
Where, $u$ object distance
$v$ is the image distance.
The negative sign represents that these coordinates are on the left side of the pole of the mirror.
Using mirror formula,
$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$
Putting the values, we get
$\dfrac{1}{{ - 2a}} + \dfrac{1}{{ - 4a}} = \dfrac{1}{f}$
$ \Rightarrow \dfrac{{ - 2 - 1}}{{4a}} = \dfrac{1}{f}$
Solving this
$\dfrac{1}{f} = \dfrac{{ - 3}}{{4a}}$
We know that, radius of curvature of a mirror is given by twice its focal length.
$R = 2f$
$ \Rightarrow R = 2\left( {\dfrac{{ - 4a}}{3}} \right)$
Putting $a = 3m$, we get
$R = 8m$.
Hence, Option (C) is correct.
Note The basic concept is to remember the mirror formulas. You should be careful if you are taking the numeric value of the image and object distance because it can easily lead you to error. A better way is to solve it first and then put the numeric value.
Recently Updated Pages
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

Charle's Law Formula - Definition, Derivation and Solved Examples

Central Angle of a Circle Formula - Definition, Theorem and FAQs

Average Force Formula - Magnitude, Solved Examples and FAQs

Boyles Law Formula - Boyles Law Equation | Examples & Definitions

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric field due to uniformly charged sphere class 12 physics JEE_Main

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Physics Average Value and RMS Value JEE Main 2025

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11
