Two objects `A `and `B` are placed at `15` cm and `25` cm from the pole in front of a concave mirror having radius of curvature `40` cm. The distance between images formed by the mirror is :

Options

A. `40 cm`
B. `160 cm`
C. `60 cm`
D. `100 cm`

Answer

To solve the problem, we need to find the distances of the images formed by a concave mirror for two objects placed at different distances from the mirror. We'll use the mirror formula:

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

where:
- \( f \) is the focal length of the mirror,
- \( v \) is the image distance,
- \( u \) is the object distance.

### Step 1: Determine the focal length of the mirror
The radius of curvature \( R \) of the concave mirror is given as \( 40 \, \text{cm} \). The focal length \( f \) is related to the radius of curvature by:

\[ f = -\frac{R}{2} = -\frac{40}{2} = -20 \, \text{cm} \]

### Step 2: Calculate the image distance for object A
For object A, the object distance \( u_1 \) is \( -15 \, \text{cm} \) (negative because it is in front of the mirror). Using the mirror formula:

\[ \frac{1}{f} = \frac{1}{v_1} + \frac{1}{u_1} \]

Substituting the known values:

\[ \frac{1}{-20} = \frac{1}{v_1} + \frac{1}{-15} \]

Rearranging gives:

\[ \frac{1}{v_1} = \frac{1}{-20} + \frac{1}{15} \]

Finding a common denominator (60):

\[ \frac{1}{v_1} = \frac{-3}{60} + \frac{4}{60} = \frac{1}{60} \]

Thus,

\[ v_1 = 60 \, \text{cm} \]

### Step 3: Calculate the image distance for object B
For object B, the object distance \( u_2 \) is \( -25 \, \text{cm} \). Again using the mirror formula:

\[ \frac{1}{f} = \frac{1}{v_2} + \frac{1}{u_2} \]

Substituting the known values:

\[ \frac{1}{-20} = \frac{1}{v_2} + \frac{1}{-25} \]

Rearranging gives:

\[ \frac{1}{v_2} = \frac{1}{-20} + \frac{1}{-25} \]

Finding a common denominator (100):

\[ \frac{1}{v_2} = \frac{-5}{100} + \frac{-4}{100} = \frac{-9}{100} \]

Thus,

\[ v_2 = -\frac{100}{9} \approx -11.11 \, \text{cm} \]

### Step 4: Calculate the distance between the images
The distance between the images \( d \) is given by the absolute difference between \( v_1 \) and \( v_2 \):

\[ d = |v_1 - v_2| = |60 - (-11.11)| = |60 + 11.11| = 71.11 \, \text{cm} \]

### Final Answer
The distance between the images formed by the mirror is approximately \( 71.11 \, \text{cm} \).

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