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A real point source is 5 cm away from a plane mirror whose reflecting ability is $50 %$ while the eye of an observer (pupil diameter $5 m m$ ) is $10 c m$ away from the mrror. Assume that both source and eye are on same line perpendicular to surface and refracted rays have no effect on intensity.

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The area of the mirror used in observing the image of source is $\frac{25 π}{36} {mm}^{2}$

The area of the mirror used in observing the image of source is $25 π {mm}^{2}$

The ratio of the intensities of light as received by observer in the presence to absence of mirror is $10 / 9$

The ratio of the intensities of light as received by observer in the presence to absence of mirror is $19 / 18 .$

answer is A, D.

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Detailed Solution

From $△ IOP &△ IBA$

$\frac{O P}{5 cm} = \frac{2.5 mm}{15 cm} \Rightarrow O P = \frac{5}{6} mm$

Area of mirror used in reflection is $= π \times {OP}^{2} = \frac{25 π}{36} {mm}^{2}$

Let power of source is P

Intensity in absence of mirror is, $I_{1} = \frac{P}{4 π \times (5)^{2}}$

Intensity in presence of mirror is

$\begin{array}{l} I_{2} = \frac{P}{4 π \times 5^{2}} + \frac{P / 2}{4 π \times 15^{2}} \\ \Rightarrow \frac{I_{2}}{I_{1}} = \frac{19}{18} \end{array}$

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