Two plane mirrors AB and AC are inclined at an angle `theta = 20^(@)`. A ray of light starting from point P is incident at point Q on the mirror AB, then at R on mirror AC and again on S on AB. Finally the ray ST goes parallel to mirror AC. The angle which the ray makes with the normal at point Q on mirror AB is

Options

A. `20^(@)`
B. `30^(@)`
C. `40^(@)`
D. `60^(@)`

Answer

B

Explanation

`90^(@)-i(theta) + 2r=180^(@)`
` :. r-((90^(@)+i-theta)/(2))`
`p=180^(@)-(90^(@)-r)-2i`
`=180^(@)-90^(@)+((90^(@)+i+theta)/(2))-2i`

` 90^(@)-r`
`=(180^(@)-4 i+ 90^(@) +i -theta)/(2)`
`=(270^(@)-3i-theta)/(2)`
`m=p-(theta)=180^(@)-(90^(@)-i)-2i`
`:. ((270^(@)-3i-(theta))/(2) -theta=90^(@)-i`
Substituting `(theta=20^(@)`, we get
`i=30^(@)`.