Question

A finite size object is placed normal to the principal axis at a distance of 30 cm from a convex mirror of focal length 30 cm. A plane mirror is now placed in such a way that the image produced by both the mirrors coincide with each other. The distance between the two mirrors is:

• A. 45 cm
• B. 7.5 cm
• C. 22.5 cm
• D. 15 cm

Hint:

The image formed by a convex mirror is virtual and behind the mirror. The distance between two mirrors can be determined by the image formed by the first mirror.

Answer 1

The Correct Option is B

To solve this problem, we need to analyze how the images from the convex and plane mirrors can coincide.

Step 1: Determine the properties of the convex mirror.

Given:

• Focal length of the convex mirror, \(f = 30 \, \text{cm}\)
• Object distance from the convex mirror, \(u = -30 \, \text{cm}\) (The negative sign indicates that the object is in front of the mirror on the principal axis)

For a convex mirror, the mirror formula is:

\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)

Substituting the given values:

\(\frac{1}{30} = \frac{1}{v} - \frac{1}{30}\)

Solving for \(v\):

\(\frac{1}{v} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30}\)

Thus, \(v = 15 \, \text{cm}\)

This means the image formed by the convex mirror is virtual, upright, and located 15 cm behind the mirror.

Step 2: Analyze the image formation by the plane mirror.

The plane mirror forms an image that appears at the same perpendicular distance behind the mirror as the object is in front of it. If the image from the convex mirror coincides with the image formed by the plane mirror, they must be at the same location.

The virtual image from the convex mirror is at 15 cm behind it. For the images to coincide:

• The distance to the plane mirror from the image point behind the convex mirror must be equal to the 15 cm image distance behind the convex mirror.

Let the distance between the two mirrors be \(D\) cm.

For the images to coincide:

\(D + D = 15 \, \text{cm}\)

Simplifying gives:

\(2D = 15 \, \text{cm}\)

\(D = 7.5 \, \text{cm}\)

Therefore, the distance between the two mirrors must be \(7.5 \, \text{cm}\).

Hence, the correct answer is 7.5 cm.

Answer 2

To solve this problem, we need to understand how the images formed by a convex mirror and a plane mirror can coincide.

1. First, let's consider the image formed by the convex mirror. The formula for the image distance (\(v\)) from the mirror is given by the mirror equation: 

\[\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\]

1. , where \(f\) is the focal length and \(u\) is the object distance.
2. Since the object distance (\(u\)) is 30 cm and it is real, it should be taken as -30 cm (conventionally, real object distances are taken as negative for mirrors). The focal length (\(f\)) of the convex mirror is +30 cm (since it is a convex mirror).
3. Substitute the given values into the mirror equation: 

\[\frac{1}{30} = \frac{1}{v} - \frac{1}{30}\]

1. .
2. Simplifying this gives: 

\[\frac{1}{v} = \frac{1}{30} + \frac{1}{30} = \frac{2}{30}\]

1. , which results in \(v = 15\) cm (image distance from the convex mirror).
2. The nature of the image formed by the convex mirror is virtual and located behind the mirror at a distance of 15 cm.
3. Now, consider the plane mirror. For the images from both mirrors to coincide, the virtual image from the convex mirror must act as a real object for the plane mirror.
4. This means the distance from the plane mirror to the coinciding image should be such that the light seems to come from this virtual image of the convex mirror. Thus, the distance between the convex and plane mirror must be \((15 - d = d)\), where \(d\) is the distance that needs to be covered.
5. Solving the equation for the distance \(d\): \(d = \frac{15}{2} = 7.5\) cm.
6. Thus, the distance between the two mirrors is 7.5 cm.

Therefore, the correct answer is 7.5 cm.