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`

Explanation

To solve the problem step by step, we need to analyze the situation involving a point source, a plane mirror, and an observer's eye. Let's break it down:

### Step 1: Understand the Setup
- A real point source (S) is located 5 cm away from a plane mirror (M).
- The observer's eye (O) is 10 cm away from the mirror.
- The mirror has a reflecting ability of 50%.

### Step 2: Determine the Image Location
- The image (I) of the point source formed by the plane mirror will be located 5 cm behind the mirror. Thus, the distance from the mirror to the image is also 5 cm.

### Step 3: Calculate the Total Distance from Source to Observer
- The total distance from the source to the observer can be calculated as follows:
- Distance from source to mirror = 5 cm
- Distance from mirror to observer = 10 cm
- Total distance (S to O) = 5 cm + 10 cm = 15 cm

### Step 4: Determine the Pupil Diameter
- The diameter of the observer's pupil is given as 5 mm, which is equivalent to 0.5 cm.

### Step 5: Set Up Similar Triangles
- We can use similar triangles to relate the distances and the diameters involved:
- Let A be the point where the light rays from the source reach the mirror.
- Let B be the point where the reflected rays reach the observer's eye.
- The triangles formed by the image and the observer's eye can be used to find the effective area of the mirror that contributes to the observer's view.

### Step 6: Calculate the Effective Area of the Mirror
- The area of the mirror that contributes to the observer's view can be calculated using the ratio of the pupil diameter to the distance from the observer to the image.
- Using the similarity of triangles:
\[ \frac{IP}{PC} = \frac{IA}{AB} \]
where:
- \( IP \) = distance from image to observer
- \( PC \) = distance from the pupil to the mirror
- \( IA \) = distance from source to image
- \( AB \) = distance from source to observer

### Step 7: Calculate the Intensity of Light
- The intensity of light received by the observer can be calculated for both cases (with and without the mirror):
- **Without the mirror**:
\[ I_1 = \frac{P}{4 \pi (5^2)} \]
- **With the mirror**:
- The intensity is halved due to the 50% reflection ability of the mirror:
\[ I_2 = \frac{P/2}{4 \pi (15^2)} \]

### Step 8: Calculate the Ratio of Intensities
- The ratio of the intensity with the mirror to the intensity without the mirror can be expressed as:
\[ \frac{I_2}{I_1} = \frac{\frac{P/2}{4 \pi (15^2)}}{\frac{P}{4 \pi (5^2)}} \]
- Simplifying this gives:
\[ \frac{I_2}{I_1} = \frac{1/2 \cdot (5^2)}{(15^2)} = \frac{25/2}{225} = \frac{25}{450} = \frac{1}{18} \]

### Final Answer
The ratio of the intensity of light received by the observer in the presence of the mirror to that in the absence of the mirror is \( \frac{19}{18} \).