A real point source is 5cm away from a plane mirror whose reflectinig ability is `50%` , while the eye of an observer (pupil diameter 5mm) is 10cm away form the mirror. Asuume that both source and eye are on the same line perpendicular to the surface and refracted rays have no effect on intensity. Then,

Options

A. the area of the mirro used in observing the image of source is `(25pi//36)mm^(2)`
B. the area of the mirror used in observing the image of source is `25pimm^(2)`
C. the ratio of the intensities of light as received by the observer in the presence to that in the absence of mirror is `(10//9)`
D. the ratio of the intensities of light as received in the presence to that in the absence of mirror is `19//18`

Answer

a.,d.

Explanation

Here, `(OP)/(5cm)=(2.5mm)/(15cm)`
`OP=(5)/(6)mm`
Area of mirror used in reflection is
`pixxOP^(2)=(25pi)/(36)mm^(2)`
Let the power of source be P.
Intensity in absence of mirror is
`I_(1)=(P)/(4pixx(5)^(2))`
Intensity in presence of mirror is
`I_(2)=(P)/(4pixx(5)^(2))+(P//2)/(4pixx(5)^2)`
`(I_(2))/(I_(1))=(19)/(18)`