Question

A plane glass mirror of thickness 3 cm of material of \( \mu = \frac{3}{2} \) is silvered on the black surface. When a point object is placed 9 cm from the front surface of the mirror, then the position of the brightest image from the front surface is:

• A. 9 cm
• B. 11 cm
• C. 12 cm
• D. 13 cm

Hint:

For mirrors with thickness, always take into account the material's refractive index for accurate distance calculations.

Answer 1

The Correct Option is D

Step 1: Understanding the situation

The problem involves a plane mirror that is silvered on its back surface, making it a "silvered mirror." The mirror has a thickness of 3 cm, and the refractive index \( \mu = \frac{3}{2} \). A point object is placed 9 cm from the front surface of the mirror. Since the mirror is silvered on the back surface, the light travels through the glass, reflects off the back surface, and then exits through the front surface. We need to find the position of the image formed by this setup.

Step 2: Effective thickness of the glass

When light enters the glass, it first travels a distance equal to the thickness of the glass (3 cm), and since the refractive index \( \mu \) is greater than 1, the apparent distance traveled by light is shortened. The effective thickness \( t_{\text{eff}} \) of the glass for light is given by: \[ t_{\text{eff}} = \frac{t}{\mu} = \frac{3 \, \text{cm}}{\frac{3}{2}} = 2 \, \text{cm} \] So, the light appears to travel only 2 cm within the glass due to the refractive index.

Step 3: Image formation by the plane mirror

For a plane mirror, the image is formed at the same distance behind the mirror as the object is in front of it, except for the additional path the light travels through the glass. The object is 9 cm from the front surface, so the effective object distance from the mirror's silvered back surface becomes: \[ 9 \, \text{cm} + 2 \, \text{cm} = 11 \, \text{cm} \] This is the distance from the back surface where the image would form (virtual image due to the silvering).

Step 4: Find the final position of the image

The image formed is behind the mirror, at a distance of 11 cm from the back surface. Since the distance from the front surface to the back surface is 3 cm, the final position of the image from the front surface is: \[ 11 \, \text{cm} + 3 \, \text{cm} = 14 \, \text{cm} \]

Final Answer:

The position of the brightest image from the front surface of the mirror is \( \boxed{14 \, \text{cm}} \).