A small object is placed on the principal axis of concave spherical mirror of radius `20 cm` at a distance of `30 cm.` By how much will the position of the image alter only after mirror, when a parallel-sided slab of glass of thickness `6 cm` and refractive index `1.5` is introduced between the centre of curvature and the object? The parallel sides are perpendicular to the principal axis.

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Determine the Focal Length of the Concave Mirror
The focal length (F) of a concave mirror is given by the formula:
\[ F = \frac{R}{2} \]
where R is the radius of curvature.

Given:
- Radius of curvature, \( R = 20 \, \text{cm} \)

Calculating the focal length:
\[ F = \frac{20 \, \text{cm}}{2} = 10 \, \text{cm} \]

### Step 2: Use the Mirror Formula to Find the Initial Image Position (V1)
The mirror formula is:
\[ \frac{1}{F} = \frac{1}{V} + \frac{1}{U} \]
where:
- \( U \) is the object distance (negative for concave mirrors),
- \( V \) is the image distance.

Given:
- Object distance, \( U = -30 \, \text{cm} \)

Substituting the values into the mirror formula:
\[ \frac{1}{10} = \frac{1}{V_1} + \frac{1}{-30} \]

Rearranging gives:
\[ \frac{1}{V_1} = \frac{1}{10} + \frac{1}{30} \]

Finding a common denominator (30):
\[ \frac{1}{V_1} = \frac{3}{30} + \frac{1}{30} = \frac{4}{30} \]
\[ V_1 = \frac{30}{4} = 7.5 \, \text{cm} \]

### Step 3: Calculate the Shift Due to the Glass Slab
The formula for the shift (d) caused by a glass slab is:
\[ d = t \left(1 - \frac{1}{\mu}\right) \]
where:
- \( t \) is the thickness of the slab,
- \( \mu \) is the refractive index.

Given:
- Thickness of the slab, \( t = 6 \, \text{cm} \)
- Refractive index, \( \mu = 1.5 \)

Calculating the shift:
\[ d = 6 \left(1 - \frac{1}{1.5}\right) = 6 \left(1 - \frac{2}{3}\right) = 6 \left(\frac{1}{3}\right) = 2 \, \text{cm} \]

### Step 4: Adjust the Object Distance (U) After Introducing the Slab
The new object distance \( U' \) after introducing the slab is:
\[ U' = U + d \]
Substituting the values:
\[ U' = -30 \, \text{cm} + 2 \, \text{cm} = -28 \, \text{cm} \]

### Step 5: Use the Mirror Formula to Find the New Image Position (V2)
Using the new object distance in the mirror formula:
\[ \frac{1}{F} = \frac{1}{V_2} + \frac{1}{U'} \]
Substituting the values:
\[ \frac{1}{10} = \frac{1}{V_2} + \frac{1}{-28} \]

Rearranging gives:
\[ \frac{1}{V_2} = \frac{1}{10} + \frac{1}{-28} \]

Finding a common denominator (140):
\[ \frac{1}{V_2} = \frac{14}{140} - \frac{5}{140} = \frac{9}{140} \]
\[ V_2 = \frac{140}{9} \approx 15.56 \, \text{cm} \]

### Step 6: Calculate the Change in Image Position
The change in image position is:
\[ \Delta V = V_2 - V_1 \]
Substituting the values:
\[ \Delta V = 15.56 \, \text{cm} - 7.5 \, \text{cm} = 8.06 \, \text{cm} \]

### Final Answer
The position of the image alters by approximately \( 8.06 \, \text{cm} \) after the introduction of the glass slab.

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