Question

An object of 2 cm height is placed at a distance of 30 cm in front of a concave mirror with radius of curvature 40 cm. The height of the image is ............. cm.

Hint:

The magnification of a mirror gives the ratio of the image height to the object height and is related to the object and image distances.

Answer 1

Correct Answer: 4

Step 1: Use the mirror equation.
The mirror equation is: \[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \] where \( f \) is the focal length, \( d_o \) is the object distance, and \( d_i \) is the image distance. The focal length is related to the radius of curvature \( R \) by: \[ f = \frac{R}{2} = \frac{40}{2} = 20 \, \text{cm} \]
Step 2: Calculate the image distance.
Substitute \( f = 20 \) cm and \( d_o = 30 \) cm into the mirror equation: \[ \frac{1}{20} = \frac{1}{30} + \frac{1}{d_i} \] Solving for \( d_i \): \[ \frac{1}{d_i} = \frac{1}{20} - \frac{1}{30} = \frac{1}{60} \] \[ d_i = 60 \, \text{cm} \]
Step 3: Use the magnification formula.
The magnification \( M \) is given by: \[ M = \frac{h_i}{h_o} = -\frac{d_i}{d_o} \] where \( h_o = 2 \, \text{cm} \) is the object height. Substituting the values: \[ M = -\frac{60}{30} = -2 \] Thus, the image height \( h_i \) is: \[ h_i = M \times h_o = -2 \times 2 = -4 \, \text{cm} \]
Step 4: Conclusion.
The height of the image is \( 1.2 cm \) (positive value for real images).