Two mirrors, one concave and the other convex, are placed 60 cm apart with their reflecting surfaces facing each other . An object is placed 30 cm from the pole of either of them on their axis. If the focal lengths of both the mirrors are 15 cm , the position of the image formed by reflection, first at the convex and then at the concave mirrors , is

Options

A. 19.09 cm from the pole of the convex mirror
B. 19.09 cm from the pole of the concave mirror
C. 11.09 cm from the pole of the concave mirror
D. 11.09 cm from the pole of the convex mirror

Answer

To solve the problem step by step, we will use the mirror formula and the sign conventions for concave and convex mirrors.

### Step 1: Identify the given data
- Distance between the two mirrors = 60 cm
- Object distance from the pole of either mirror, \( u = -30 \) cm (negative because the object is in front of the mirror)
- Focal length of the convex mirror, \( f_1 = +15 \) cm (positive for convex mirrors)
- Focal length of the concave mirror, \( f_2 = -15 \) cm (negative for concave mirrors)

### Step 2: Find the image distance for the convex mirror
Using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
For the convex mirror:
\[ \frac{1}{f_1} = \frac{1}{v_1} + \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{15} = \frac{1}{v_1} - \frac{1}{30} \]
Rearranging gives:
\[ \frac{1}{v_1} = \frac{1}{15} + \frac{1}{30} \]
Finding a common denominator (30):
\[ \frac{1}{v_1} = \frac{2}{30} + \frac{1}{30} = \frac{3}{30} = \frac{1}{10} \]
Thus,
\[ v_1 = 10 \text{ cm} \]

### Step 3: Determine the object distance for the concave mirror
The image formed by the convex mirror acts as the object for the concave mirror. Since the mirrors are 60 cm apart, the object distance for the concave mirror, \( u_2 \), is:
\[ u_2 = - (60 - v_1) = - (60 - 10) = -70 \text{ cm} \]

### Step 4: Find the image distance for the concave mirror
Using the mirror formula again for the concave mirror:
\[ \frac{1}{f_2} = \frac{1}{v_2} + \frac{1}{u_2} \]
Substituting the values:
\[ \frac{1}{-15} = \frac{1}{v_2} - \frac{1}{70} \]
Rearranging gives:
\[ \frac{1}{v_2} = \frac{1}{-15} + \frac{1}{70} \]
Finding a common denominator (210):
\[ \frac{1}{v_2} = \frac{-14}{210} + \frac{3}{210} = \frac{-11}{210} \]
Thus,
\[ v_2 = -\frac{210}{11} \approx -19.09 \text{ cm} \]

### Step 5: Conclusion
The final image formed by the concave mirror is located at approximately 19.09 cm from the pole of the concave mirror, and since it is negative, it indicates that the image is formed on the same side as the object.

### Final Answer
The position of the image formed by reflection first at the convex and then at the concave mirror is approximately **19.09 cm from the pole of the concave mirror**.

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