
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
430.6k+ 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
Young's Double Slit Experiment Step by Step Derivation

How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Chemical Equation - Important Concepts and Tips for JEE

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

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Physics Average Value and RMS Value JEE Main 2025

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

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

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

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Collision - Important Concepts and Tips for JEE

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

Charging and Discharging of Capacitor
