An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of film?

Options

A. 7.2 m
B. 2.4 m
C. 3.2 m
D. 5.6 m

Answer

To solve the problem step-by-step, we will follow the given information and apply the lens formula, along with the concept of the shift caused by the glass plate.

### Step 1: Identify the given values
- Object distance (u) = -2.4 m (since object distance is taken as negative in lens formula)
- Image distance (v) = 12 cm = 0.12 m (positive since it is real and formed on the opposite side of the lens)
- Thickness of the glass plate (t) = 1 cm = 0.01 m
- Refractive index of the glass plate (n) = 1.50

### Step 2: Calculate the focal length (f) of the lens using the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the values:
\[ \frac{1}{f} = \frac{1}{0.12} - \frac{1}{-2.4} \]
Calculating each term:
\[ \frac{1}{0.12} = \frac{100}{12} = \frac{25}{3} \]
\[ \frac{1}{-2.4} = -\frac{1}{2.4} = -\frac{5}{12} \]
Now substituting back:
\[ \frac{1}{f} = \frac{25}{3} + \frac{5}{12} \]
Finding a common denominator (12):
\[ \frac{1}{f} = \frac{100}{12} + \frac{5}{12} = \frac{105}{12} \]
Thus,
\[ f = \frac{12}{105} = \frac{4}{35} \text{ m} \approx 0.1143 \text{ m} \text{ or } 11.43 \text{ cm} \]

### Step 3: Calculate the shift caused by the glass plate
The shift (Δ) caused by the glass plate can be calculated using the formula:
\[ \Delta = t \left(1 - \frac{1}{n}\right) \]
Substituting the values:
\[ \Delta = 0.01 \left(1 - \frac{1}{1.50}\right) = 0.01 \left(1 - \frac{2}{3}\right) = 0.01 \left(\frac{1}{3}\right) = \frac{0.01}{3} \approx 0.00333 \text{ m} \text{ or } 0.333 \text{ cm} \]

### Step 4: Calculate the new image distance (v')
The new image distance (v') will be:
\[ v' = v - \Delta = 0.12 - 0.00333 \approx 0.11667 \text{ m} \text{ or } 11.67 \text{ cm} \]

### Step 5: Calculate the new object distance (u')
Using the lens formula again with the new image distance:
\[ \frac{1}{f} = \frac{1}{v'} - \frac{1}{u'} \]
Rearranging gives:
\[ \frac{1}{u'} = \frac{1}{v'} - \frac{1}{f} \]
Substituting the values:
\[ \frac{1}{u'} = \frac{1}{0.11667} - \frac{105}{12} \]
Calculating each term:
\[ \frac{1}{0.11667} \approx 8.571 \text{ (approximately)} \]
Now substituting:
\[ \frac{1}{u'} = 8.571 - 8.333 = 0.238 \]
Thus,
\[ u' \approx \frac{1}{0.238} \approx 4.20 \text{ m} \]

### Step 6: Conclusion
The object should be shifted to approximately 4.20 m from the lens to form a sharp focus on the film.