What will be the number of images seen when two plane mirrors are placed at an angle of 36°

Text Solution

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

Solve Numerical example.

Estimate the number of images produced if a tiny object is kept in between two plane mirrors inclined at 35°, 36°, 40° and 45°.

1. For θ1 = 35°
   n1 = `360/θ_1=360/35` = 10.28
   As n1 is non-integer, N1 = integral part of n1 = 10
2. For θ2 = 36°
   n2 = `360/36` = 10
   As n2 is even integer, N2 = (n2 – 1) = 9
3. or θ3 = 40°
   n3 = `360/40` = 9
   As n3 is odd integer,
   Number of images seen (N3) = n3 – 1= 8(if the object is placed at the angle bisector)

   or Number of images seen (N3) = n3 = 9

   
   (if the object is placed off the angle bisector)
4. For θ4 = 45°
   n4 = `360/45` = 8
   As n4 is even integer,
   N4 = n4 – 1 = 7

The number of images seen when mirrors are inclined at 35°, 36°, 40°, 45° are 10, 9, 8 or 9, 7 respectively.

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