A lens with a focal length of `f=30 cm` produces on a screen a sharp image of an object that is at a distance of `a=40 cm` from the lens. A plane parallel plate with thickness of `d=9 cm` is placed between the lens and the object perpendicular to the optical axis of the lens. Through what distance should the screen be shifted for the image of the object to remain distinct? The refractive index of the glass of the plate is `mu=1.8.`

Explanation

To solve the problem step by step, we will use the lens formula and the concept of refraction through a parallel plate.

### Step 1: Identify the given values
- Focal length of the lens, \( f = 30 \, \text{cm} \)
- Object distance, \( a = 40 \, \text{cm} \) (which will be taken as \( u = -40 \, \text{cm} \) for the lens formula)
- Thickness of the plate, \( d = 9 \, \text{cm} \)
- Refractive index of the glass, \( \mu = 1.8 \)

### Step 2: Calculate the initial image distance using the lens formula
The lens formula is given by:
\[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
Substituting the known values:
\[ \frac{1}{30} = \frac{1}{v} - \frac{1}{-40} \]
This simplifies to:
\[ \frac{1}{30} = \frac{1}{v} + \frac{1}{40} \]
To solve for \( v \), we first find a common denominator:
\[ \frac{1}{30} = \frac{1}{v} + \frac{1}{40} \implies \frac{1}{v} = \frac{1}{30} - \frac{1}{40} \]
Calculating the right side:
\[ \frac{1}{30} = \frac{4}{120}, \quad \frac{1}{40} = \frac{3}{120} \implies \frac{1}{v} = \frac{4}{120} - \frac{3}{120} = \frac{1}{120} \]
Thus,
\[ v = 120 \, \text{cm} \]

### Step 3: Calculate the shift due to the parallel plate
The formula for the effective shift \( \Delta x \) caused by the parallel plate is:
\[ \Delta x = t \left(1 - \frac{1}{\mu}\right) \]
Substituting the values:
\[ \Delta x = 9 \left(1 - \frac{1}{1.8}\right) = 9 \left(1 - 0.5556\right) = 9 \times 0.4444 \approx 4 \, \text{cm} \]

### Step 4: Calculate the new object distance
The new object distance \( u' \) after placing the plate is:
\[ u' = u + \Delta x = -40 + 4 = -36 \, \text{cm} \]

### Step 5: Calculate the new image distance using the lens formula again
Using the lens formula again:
\[ \frac{1}{f} = \frac{1}{v'} - \frac{1}{u'} \]
Substituting the known values:
\[ \frac{1}{30} = \frac{1}{v'} - \frac{1}{-36} \]
This simplifies to:
\[ \frac{1}{30} = \frac{1}{v'} + \frac{1}{36} \]
Finding a common denominator:
\[ \frac{1}{v'} = \frac{1}{30} - \frac{1}{36} \]
Calculating the right side:
\[ \frac{1}{30} = \frac{6}{180}, \quad \frac{1}{36} = \frac{5}{180} \implies \frac{1}{v'} = \frac{6}{180} - \frac{5}{180} = \frac{1}{180} \]
Thus,
\[ v' = 180 \, \text{cm} \]

### Step 6: Calculate the distance the screen needs to be shifted
The distance the screen should be shifted \( \Delta s \) is:
\[ \Delta s = v' - v = 180 - 120 = 60 \, \text{cm} \]

### Final Answer
The screen should be shifted by **60 cm** to keep the image distinct.

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