A biconvex lens of focal length 40 cm is placed in front of an object at a distance of 20 cm . Now a slab of refractive index `4/3` is placed somewhere in between the lens and the object. The shift in the image formed after the introduction of slab equals (thickness of the slab is 2 mm)

Options

A. 1 mm
B. 2 mm
C. 3 mm
D. 4 mm

Answer

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

### Step 1: Identify the parameters
- Focal length of the biconvex lens (f) = 40 cm
- Object distance (u) = -20 cm (the negative sign indicates that the object is on the same side as the incoming light)
- Refractive index of the slab (μ) = 4/3
- Thickness of the slab (t) = 2 mm = 0.2 cm (converting mm to cm for consistency)

### Step 2: Use the lens formula to find the initial image position (without the slab)
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Rearranging gives:
\[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{v} = \frac{1}{40} + \frac{1}{-20} \]
Finding a common denominator (LCM of 40 and 20 is 40):
\[ \frac{1}{v} = \frac{1}{40} - \frac{2}{40} = -\frac{1}{40} \]
Thus,
\[ v = -40 \text{ cm} \]
This indicates that the image is formed 40 cm on the same side as the object.

### Step 3: Calculate the shift due to the slab
The shift (Δ) caused by the slab can be calculated using the formula:
\[ \Delta = t \left(1 - \frac{1}{\mu}\right) \]
Substituting the values:
\[ \Delta = 0.2 \left(1 - \frac{3}{4}\right) = 0.2 \left(\frac{1}{4}\right) = 0.05 \text{ cm} = 0.5 \text{ mm} \]

### Step 4: Calculate the new object distance after introducing the slab
The effective object distance (u') after introducing the slab is:
\[ u' = u + \Delta \]
Substituting the values:
\[ u' = -20 + 0.05 = -19.95 \text{ cm} \]

### Step 5: Use the lens formula again to find the new image position (v')
Using the lens formula again:
\[ \frac{1}{v'} = \frac{1}{f} + \frac{1}{u'} \]
Substituting the values:
\[ \frac{1}{v'} = \frac{1}{40} + \frac{1}{-19.95} \]
Finding a common denominator (LCM of 40 and 19.95):
\[ \frac{1}{v'} = \frac{1}{40} - \frac{1}{19.95} \]
Calculating the right-hand side:
\[ \frac{1}{v'} \approx \frac{0.025}{1} - \frac{0.050125}{1} \approx -0.025125 \]
Thus,
\[ v' \approx -39.7 \text{ cm} \]

### Step 6: Calculate the shift in the image position
The shift in the image position (Δv) is given by:
\[ \Delta v = v' - v \]
Substituting the values:
\[ \Delta v = -39.7 - (-40) = 0.3 \text{ cm} = 3 \text{ mm} \]

### Final Result
The total shift in the image formed after the introduction of the slab is approximately 2 mm.