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°.
The number of images seen when mirrors are inclined at 35°, 36°, 40°, 45° are 10, 9, 8 or 9, 7 respectively. Is there an error in this question or solution? Answer Vedantu Improvement Promise |