A convex mirror and a concave mirror of radius 10cm each are placed 15cm apart facing each other. An object is placed midway between them. If the reflection first takes place in the concave mirror and then in convex mirror, the position of the final image is

Options

A. on the pole of the convex mirror
B. on the pole of the concave mirror
C. at a distance of 10cm from the convex mirror
D. at a distance of 5cm from the concave mirror

Explanation

To solve the problem step by step, we will follow the sequence of reflections as described in the question.

### Step 1: Understand the Setup
We have a concave mirror and a convex mirror, both with a radius of curvature \( R = 10 \, \text{cm} \). The focal lengths (\( f \)) of the mirrors can be calculated as:
- For the concave mirror: \( f = -\frac{R}{2} = -\frac{10}{2} = -5 \, \text{cm} \)
- For the convex mirror: \( f = \frac{R}{2} = \frac{10}{2} = 5 \, \text{cm} \)

The distance between the two mirrors is \( 15 \, \text{cm} \), and the object is placed midway, which means it is \( 7.5 \, \text{cm} \) from each mirror.

### Step 2: Reflection in the Concave Mirror
The object distance (\( u \)) for the concave mirror is:
\[ u = -7.5 \, \text{cm} \quad (\text{negative because it is in front of the mirror}) \]

Using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{-5} = \frac{1}{v} + \frac{1}{-7.5} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{-5} + \frac{1}{7.5} \]
Finding a common denominator (which is 15):
\[ \frac{1}{v} = \frac{-3}{15} + \frac{2}{15} = \frac{-1}{15} \]
Thus,
\[ v = -15 \, \text{cm} \]
This indicates that the image formed by the concave mirror is virtual and located \( 15 \, \text{cm} \) behind the mirror.

### Step 3: Use the Image from the Concave Mirror as the Object for the Convex Mirror
The distance of this virtual image from the convex mirror is:
\[ u' = 15 \, \text{cm} + 7.5 \, \text{cm} = 22.5 \, \text{cm} \quad (\text{positive because it is in front of the convex mirror}) \]

Now, we will use this object distance for the convex mirror:
\[ u' = -22.5 \, \text{cm} \]

Using the mirror formula for the convex mirror:
\[ \frac{1}{f} = \frac{1}{v'} + \frac{1}{u'} \]
Substituting the values:
\[ \frac{1}{5} = \frac{1}{v'} + \frac{1}{-22.5} \]
Rearranging gives:
\[ \frac{1}{v'} = \frac{1}{5} + \frac{1}{22.5} \]
Finding a common denominator (which is 22.5):
\[ \frac{1}{v'} = \frac{4.5}{22.5} + \frac{1}{22.5} = \frac{5.5}{22.5} \]
Thus,
\[ v' = \frac{22.5}{5.5} \approx 4.09 \, \text{cm} \]

### Final Position of the Image
The final image formed by the convex mirror is approximately \( 4.09 \, \text{cm} \) in front of the convex mirror.

### Summary of Steps
1. Calculate the focal lengths of both mirrors.
2. Determine the object distance for the concave mirror.
3. Use the mirror formula to find the image distance from the concave mirror.
4. Calculate the new object distance for the convex mirror using the image from the concave mirror.
5. Use the mirror formula again to find the final image distance from the convex mirror.