To solve the problem step by step, we will follow these calculations:
### Step 1: Identify the Given Values
- Focal length of the lens, \( F = 40 \, \text{cm} \)
- Object distance, \( U = -20 \, \text{cm} \) (negative because it is on the opposite side of the light direction)
- Refractive index of the slab, \( \mu = \frac{4}{3} \)
- Thickness of the slab, \( t = 2 \, \text{mm} = 0.2 \, \text{cm} \)
### Step 2: Use the Lens Formula to Find the Initial Image Distance
The lens formula is given by:
\[
\frac{1}{V} - \frac{1}{U} = \frac{1}{F}
\]
Substituting the values:
\[
\frac{1}{V} - \frac{1}{-20} = \frac{1}{40}
\]
This simplifies to:
\[
\frac{1}{V} + \frac{1}{20} = \frac{1}{40}
\]
Now, finding a common denominator (which is 40):
\[
\frac{1}{V} + \frac{2}{40} = \frac{1}{40}
\]
Subtracting \( \frac{2}{40} \) from both sides:
\[
\frac{1}{V} = \frac{1}{40} - \frac{2}{40} = -\frac{1}{40}
\]
Thus, we find:
\[
V = -40 \, \text{cm}
\]
This indicates that the image is formed 40 cm on the same side as the object.
### Step 3: Calculate the Shift in Object Distance Due to the Slab
The shift \( \Delta U \) due to the slab is given by the formula:
\[
\Delta U = t \left(1 - \frac{1}{\mu}\right)
\]
Substituting the values:
\[
\Delta U = 0.2 \left(1 - \frac{1}{\frac{4}{3}}\right) = 0.2 \left(1 - \frac{3}{4}\right) = 0.2 \left(\frac{1}{4}\right)
\]
Calculating this gives:
\[
\Delta U = 0.2 \times 0.25 = 0.05 \, \text{cm} = 0.5 \, \text{mm}
\]
### Step 4: Use the Lens Formula Again to Find the New Image Distance
Now, the new object distance \( U' \) will be:
\[
U' = U + \Delta U = -20 + 0.05 = -19.95 \, \text{cm}
\]
Using the lens formula again:
\[
\frac{1}{V'} - \frac{1}{U'} = \frac{1}{F}
\]
Substituting the new object distance:
\[
\frac{1}{V'} - \frac{1}{-19.95} = \frac{1}{40}
\]
This simplifies to:
\[
\frac{1}{V'} + \frac{1}{19.95} = \frac{1}{40}
\]
Finding a common denominator (which is 40):
\[
\frac{1}{V'} + \frac{2}{39.8} = \frac{1}{40}
\]
Now, we can solve for \( V' \).
### Step 5: Calculate the Change in Image Distance
Using the differentiation technique:
\[
\frac{dV}{dU} = \frac{V^2}{U^2}
\]
Substituting the values:
\[
dV = \frac{(-40)^2}{(-20)^2} \times \Delta U = \frac{1600}{400} \times 0.05 = 4 \times 0.05 = 0.2 \, \text{cm} = 2 \, \text{mm}
\]
### Final Result
The shift in the image formed after the introduction of the slab is:
\[
\text{Shift in image} = 2 \, \text{mm}
\]
