Consider two plane mirrors inclined at an angle `theta` as shown in figure. Find the number of object O formed these mirror when

(i)`theta=60^(@)`
(ii) `theta=72^(@)` and the object at `40^(@)` from `M_(1)`
(iii) `theta=72^(@)` and the object is at `36^(@)` from `M_(1)`
(iv) `theta=80^(@)`

Explanation

Two mirrors `M_(1)` and `M_(2)` are inclined at an angle `theta`.
Let,N=`(360^(@))/(theta)`
(i) For `theta=60^(@),N=(360^(@))/(60^(@))=6`=even integer
`therefore` Number of images formed
n=N-1=6-1=5
(ii) For `theta=72^(@),N=(360^(@))/(72^(@))=5=`odd integer and the object is not on the bisector,
`therefore` Number of images formed ,n=N=5
(iii) For `theta=72^(@),N=(360^(@))/(72^(@))=5=`odd integer also bisector is at `36^(@)` and not at `30^(@)`.
`therefore` Number of images formed
n=N-1=5-1=4
(iv) For `theta=80^(@),N=(360^(@))/(80^(@))=4.5=`fraction number
`therefore` Number of images formed,
n=integral part=4.