What is the power of a concave mirror if it produces a real image twice the height of the object placed at 30 cm from the mirror?
A) \[5D\]
B) \[ - 5D\]
C) \[0.05D\]
D) \[ - 0.05D\]
Answer
270k+ views
Hint: In this solution, we will use the concepts of magnification and the mirror formula to determine the focal length of the mirror. The power of a mirror is the inverse of its focal length. We will use these relations to determine the power of the concave mirror.
Formula used: In this solution we will be using the following formula,
Mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ where $v$ is the image distance, $u$ is the object distance and $f$ is the focal length of the mirror
Magnification of a mirror: $m = \dfrac{v}{u}$
Complete step by step answer:
We’ve been given that the concave mirror produces an image twice the height of the object. So, we can say that
$m = \dfrac{v}{u} = 2$
Which gives us $v = 2u$
Now since $u = - 30\,cm$ (assuming it is placed on the left of the mirror), we will get
$v = - 60\,cm$
Using the mirror formula, we can write that
$\dfrac{1}{{ - 60}} + \dfrac{1}{{ - 30}} = \dfrac{1}{f}$
Solving for the focal length, we get
$\dfrac{1}{f} = \dfrac{3}{{ - 60}}\,cm$
On taking inverse
$f = - 20\,cm$
Now the power of a mirror is the inverse of its focal length. The focal length of the mirror in question is $f = - 20\,cm$.
This means the power will be
$P = - 1/20$
$ \Rightarrow P = - 0.05\,D$ which corresponds to option (D).
Note: While using the mirror formula, we should be careful in taking the sign convention of the object image and the focal length of the system. By convention, the object is usually assumed to be placed left of the mirror as we have done in the question. So, the reflective surface of the mirror should also be facing the object which means for a concave lens, the focal length will also be negative which can help us in narrowing the choices down to options (B) and (D).
Formula used: In this solution we will be using the following formula,
Mirror formula: $\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$ where $v$ is the image distance, $u$ is the object distance and $f$ is the focal length of the mirror
Magnification of a mirror: $m = \dfrac{v}{u}$
Complete step by step answer:
We’ve been given that the concave mirror produces an image twice the height of the object. So, we can say that
$m = \dfrac{v}{u} = 2$
Which gives us $v = 2u$
Now since $u = - 30\,cm$ (assuming it is placed on the left of the mirror), we will get
$v = - 60\,cm$
Using the mirror formula, we can write that
$\dfrac{1}{{ - 60}} + \dfrac{1}{{ - 30}} = \dfrac{1}{f}$
Solving for the focal length, we get
$\dfrac{1}{f} = \dfrac{3}{{ - 60}}\,cm$
On taking inverse
$f = - 20\,cm$
Now the power of a mirror is the inverse of its focal length. The focal length of the mirror in question is $f = - 20\,cm$.
This means the power will be
$P = - 1/20$
$ \Rightarrow P = - 0.05\,D$ which corresponds to option (D).
Note: While using the mirror formula, we should be careful in taking the sign convention of the object image and the focal length of the system. By convention, the object is usually assumed to be placed left of the mirror as we have done in the question. So, the reflective surface of the mirror should also be facing the object which means for a concave lens, the focal length will also be negative which can help us in narrowing the choices down to options (B) and (D).
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Kinematics Mock Test for JEE Main 2025-26: Comprehensive Practice

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

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

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced 2026 Marks vs Rank: Estimate IIT Rank from Your Score

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

