VerifiedHint: When a ray of light travels through a glass prism, it undergoes refraction and gets deviated from its original path. The deviation produced by a small angled prism is always a constant. Additionally, it is given when the mirror is rotated by X degrees then the total deviation of the ray becomes ${90^ \circ }C$. Substitute these values to the basic formal and solve for the answer.
Complete answer:
When light ray enters through a glass prism, the emergent ray is not parallel to the incident ray after refraction. Rather, the emergent ray deviates from its original direction by a certain angle, known as the angle of deviation.
$ {\text{In case of a prism the deviation, }}{\delta _m}{\text{ of the emergent ray is given by:}} $
$ \mu {\text{ = }}\dfrac{{\dfrac{{A + {\delta _m}}}{2}}}{{\sin \dfrac{A}{2}}} $
$ {\text{If the angle of prism A is small,}} $
$ {\delta _m}{\text{ is also small}}{\text{. So the equation becomes:}} $
$ {\delta _m} = \left( {\mu - 1} \right)A $
So, the deviation produced via a small angled prism is always, given by
$ {\delta _1} = \left( {\mu - 1} \right)\alpha = \left( {1.5 - 1} \right){4^ \circ } $
$ {\delta _1} = {2^ \circ }{\text{ }}\left[ {Always} \right] $
Deviation caused by mirror will be:
$ {\delta _2} = {180^ \circ } - 2i $
$ {\delta _2} = {180^ \circ } - 2 \times {45^ \circ } $
$ {\delta _2} = {90^ \circ } $
Thus, the net deviation produced by the system will be
$ {\delta _1} + {\delta _2} = {2^ \circ } + {90^ \circ } $
$ {\delta _1} + {\delta _2} = {92^ \circ } $
Clearly, the total deviation is more than ${90^ \circ }$.
If the angle of incidence on the mirror is greater than its associated deviation will be smaller. Let X be the angle of rotation of mirror in clockwise direction done to increase the angle of incidence, so deviation produced by the mirror will be:
${180^ \circ } - 2\left( {{{45}^ \circ } + X} \right) = {90^ \circ } - 2X$
Hence, total deviation produced ${90^ \circ } - 2X + {2^ \circ } = {92^ \circ } - 2X$
But,
$ {92^ \circ } - 2X = {90^ \circ }{\text{ }}\left[ {given} \right] $
$ \Rightarrow X = {1^ \circ } $
Therefore, the mirror is rotated by ${1^ \circ }$ degree then the total deviation of ray becomes ${90^ \circ }C$.
Note: Draw a well-labeled diagram of the given reflection and refraction scenario for a better understanding of the given question as visual cues will help. Formulas and universal facts like the deviation produced by a small angled prism are always a constant must be learned by the students beforehand.
