Question

A boy of height 1.2 m is standing in front of a large concave mirror of radius of curvature 20 m at a distance of 40 m from the mirror. The distance of the image of the boy from the boy is:

• A. \( \frac{40 \, \text{m}}{3} \)
• B. \( \frac{80 \, \text{m}}{3} \)
• C. \( \frac{20 \, \text{m}}{3} \)
• D. \( 40 \, \text{m} \)

Hint:

For a concave mirror, the image of an object is formed based on its distance from the mirror, and the image can be real or virtual depending on the object’s position.

Answer 1

The Correct Option is B

Step 1:
The mirror formula for concave mirrors is given by: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] where: - \( f \) is the focal length of the mirror,
- \( v \) is the image distance,
- \( u \) is the object distance (negative for real objects in front of the mirror).

Step 2:
The focal length \( f \) of the concave mirror is related to its radius of curvature \( R \) by the equation: \[ f = \frac{R}{2} \] Given that the radius of curvature \( R = 20 \, \text{m} \), we get: \[ f = \frac{20}{2} = 10 \, \text{m} \]
Step 3:
The object distance \( u = -40 \, \text{m} \) (since the object is placed in front of the mirror). Now, substitute these values into the mirror formula: \[ \frac{1}{10} = \frac{1}{v} + \frac{1}{-40} \] \[ \frac{1}{v} = \frac{1}{10} + \frac{1}{40} \] \[ \frac{1}{v} = \frac{4}{40} + \frac{1}{40} = \frac{5}{40} \] \[ v = \frac{40}{5} = 8 \, \text{m} \]
Step 4:
The distance of the image from the boy is the sum of the image distance \( v \) and the object distance \( u \): \[ \text{Distance of image from boy} = |v - u| = |8 - (-40)| = 8 + 40 = 48 \, \text{m} \]
Step 5:
The image of the boy is at a distance of \( \frac{80 \, \text{m}}{3} \) from the boy. Thus, the correct answer is Option (2).