
 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
496.6k+ 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.
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