Formula for number of images formed by two plane mirrors incident at an angle $\theta $ is $n$ = $\dfrac{{360^\circ }}{\theta }$. If n is even, the number of images is n-1, if n is an odd number of images.
Column I Column II a) $\theta $ $ = $ $60^\circ $ 1) n $ = $ 9 b) $\theta $ $ = $ $60^\circ $ 2) n $ = $ 3 c) $\theta $ $ = $ $60^\circ $ 3) n $ = $ 5 d) $\theta $ $ = $ $60^\circ $ 4) n $ = $ 7 5) n $ = $ 1
| Column I | Column II |
| a) $\theta $ $ = $ $60^\circ $ | 1) n $ = $ 9 |
| b) $\theta $ $ = $ $60^\circ $ | 2) n $ = $ 3 |
| c) $\theta $ $ = $ $60^\circ $ | 3) n $ = $ 5 |
| d) $\theta $ $ = $ $60^\circ $ | 4) n $ = $ 7 |
| 5) n $ = $ 1 |
Answer
575.5k+ views
Hint: Image is defined as the collection of focus points of light rays coming from an object. If the image of the object is viewed in two plane mirrors that are inclined to each other, more than one image is formed. The number of images formed by two plane mirrors depends on the angle between the mirror.
Complete step by step solution:
Given the angle is $\theta $.
If the value of $\dfrac{{360^\circ }}{\theta }$is even, then we will use the formula
No. of images $ = $$\dfrac{{360^\circ }}{\theta } - 1$
If the value $\dfrac{{360^\circ }}{\theta }$is odd, then we will use the formula
No. of images $ = $$\dfrac{{360^\circ }}{\theta }$
a) When $\theta $ $ = $ $60^\circ $
Let us find the value of $\dfrac{{360^\circ }}{\theta }$
So, $\dfrac{{360^\circ }}{{60^\circ }}$$ = $ 6, where 6 is an even number.
we will use the formula for No. of images $ = $ $\dfrac{{360^\circ }}{\theta } - 1$
$ \Rightarrow $ 6 $ - $1 $ = $ 5
Thus, the images formed will be 5.
b) When $\theta $ $ = $ $40^\circ $
Let us find the value of $\dfrac{{360^\circ }}{\theta }$
So, $\dfrac{{360^\circ }}{{40^\circ }}$$ = $ 9, where 9 is an odd number.
we will use the formula for No. of images $ = $$\dfrac{{360^\circ }}{\theta }$$ = $ 9
Thus, the images formed will be 9.
c) When $\theta $ $ = $ $90^\circ $
Let us find the value of $\dfrac{{360^\circ }}{\theta }$
So, $\dfrac{{360^\circ }}{{90^\circ }}$$ = $ 4, where 4 is an even number.
we will use the formula for No. of images $ = $$\dfrac{{360^\circ }}{\theta } - 1$
$ \Rightarrow $ 4 $ - $1 $ = $ 3
Thus, the images formed will be 3.
d) When $\theta $ $ = $ $180^\circ $
Let us find the value of $\dfrac{{360^\circ }}{\theta }$
So, $\dfrac{{360^\circ }}{{180^\circ }}$$ = $ 2, where 2 is an even number.
we will use the formula for No. of images $ = $$\dfrac{{360^\circ }}{\theta } - 1$
$ \Rightarrow $ 2 $ - $1 $ = $ 1
Thus, the images formed will be 1.
Hence the correct option for the problem is a $ = $3, b $ = $1, c $ = $2, d $ = $5.
Note: 1) If $\dfrac{{360^\circ }}{\theta }$ is a fraction, then the number of images formed will be equal to its integral part.
2) The smaller the angle, the greater the number of images.
Complete step by step solution:
Given the angle is $\theta $.
If the value of $\dfrac{{360^\circ }}{\theta }$is even, then we will use the formula
No. of images $ = $$\dfrac{{360^\circ }}{\theta } - 1$
If the value $\dfrac{{360^\circ }}{\theta }$is odd, then we will use the formula
No. of images $ = $$\dfrac{{360^\circ }}{\theta }$
a) When $\theta $ $ = $ $60^\circ $
Let us find the value of $\dfrac{{360^\circ }}{\theta }$
So, $\dfrac{{360^\circ }}{{60^\circ }}$$ = $ 6, where 6 is an even number.
we will use the formula for No. of images $ = $ $\dfrac{{360^\circ }}{\theta } - 1$
$ \Rightarrow $ 6 $ - $1 $ = $ 5
Thus, the images formed will be 5.
b) When $\theta $ $ = $ $40^\circ $
Let us find the value of $\dfrac{{360^\circ }}{\theta }$
So, $\dfrac{{360^\circ }}{{40^\circ }}$$ = $ 9, where 9 is an odd number.
we will use the formula for No. of images $ = $$\dfrac{{360^\circ }}{\theta }$$ = $ 9
Thus, the images formed will be 9.
c) When $\theta $ $ = $ $90^\circ $
Let us find the value of $\dfrac{{360^\circ }}{\theta }$
So, $\dfrac{{360^\circ }}{{90^\circ }}$$ = $ 4, where 4 is an even number.
we will use the formula for No. of images $ = $$\dfrac{{360^\circ }}{\theta } - 1$
$ \Rightarrow $ 4 $ - $1 $ = $ 3
Thus, the images formed will be 3.
d) When $\theta $ $ = $ $180^\circ $
Let us find the value of $\dfrac{{360^\circ }}{\theta }$
So, $\dfrac{{360^\circ }}{{180^\circ }}$$ = $ 2, where 2 is an even number.
we will use the formula for No. of images $ = $$\dfrac{{360^\circ }}{\theta } - 1$
$ \Rightarrow $ 2 $ - $1 $ = $ 1
Thus, the images formed will be 1.
Hence the correct option for the problem is a $ = $3, b $ = $1, c $ = $2, d $ = $5.
Note: 1) If $\dfrac{{360^\circ }}{\theta }$ is a fraction, then the number of images formed will be equal to its integral part.
2) The smaller the angle, the greater the number of images.
Recently Updated Pages
Properties of Solids and Liquids Mock Test 2025

JEE Main Mock Test 2025-26: Dual Nature of Matter & Radiation

JEE Main 2025-26 Work, Energy and Power Mock Test – Free Practice Online

JEE Main Mock Test 2025-26: Experimental Skills Chapter Online Practice

JEE Main 2025-26 Mock Test: Properties of Solids and Liquids

JEE Main 2025 Kinetic Theory Of Gases Mock Test: Practice & Solutions

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

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

Electron Gain Enthalpy and Electron Affinity Explained

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

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

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

Understanding Uniform Acceleration in Physics

Understanding Electromagnetic Waves and Their Importance

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

