The objective and eye piece of an astronomical telescope are double convexlenses with refractive jndex 1.5. When the telescope is adjusted to infinity the seperation between the lenses is 16 cm. If the space between the lenses is now filled with water and again telescope is adjusted for infinity. Then the present separation between the lenses is

Options

A. 8 cm
B. 16 cm
C. 24 cm
D. 32 cm

Answer

To solve the problem, we need to determine the new separation between the objective and eyepiece lenses of an astronomical telescope after the space between them is filled with water. Here's the step-by-step solution:

### Step 1: Understand the initial conditions
The initial separation between the lenses when adjusted for infinity is given as 16 cm. We denote this separation as \( S_0 = 16 \, \text{cm} \).

### Step 2: Refractive indices
The refractive index of the lenses (made of glass) is given as \( \mu_1 = 1.5 \). The refractive index of water is \( \mu_2 = 1.33 \). The refractive index of air is \( \mu_3 = 1.0 \).

### Step 3: Use lens maker's formula
When the telescope is adjusted to infinity, the effective focal length \( F \) of the telescope can be expressed in terms of the focal lengths of the objective \( f_o \) and the eyepiece \( f_e \):
\[ \frac{1}{F} = \frac{1}{f_o} + \frac{1}{f_e} \]
The total length of the telescope is given by:
\[ S_0 = f_o + f_e \]

### Step 4: Calculate the focal lengths
From the lens maker's formula, we know that the focal length \( f \) of a lens is given by:
\[ f = \frac{R}{(\mu - 1)} \]
where \( R \) is the radius of curvature of the lens and \( \mu \) is the refractive index.

For the objective lens:
\[ f_o = \frac{R_o}{(1.5 - 1)} = \frac{R_o}{0.5} = 2R_o \]

For the eyepiece lens:
\[ f_e = \frac{R_e}{(1.5 - 1)} = \frac{R_e}{0.5} = 2R_e \]

### Step 5: Substitute into the total length equation
Substituting the focal lengths into the total length equation:
\[ S_0 = f_o + f_e = 2R_o + 2R_e = 2(R_o + R_e) \]
Since \( S_0 = 16 \, \text{cm} \):
\[ R_o + R_e = 8 \, \text{cm} \]

### Step 6: Adjust for water
When the space between the lenses is filled with water, we need to recalculate the effective focal lengths using the refractive index of water. The new focal lengths become:
\[ f_o' = \frac{R_o}{(1.33 - 1)} = \frac{R_o}{0.33} \]
\[ f_e' = \frac{R_e}{(1.33 - 1)} = \frac{R_e}{0.33} \]

### Step 7: Calculate new total length
The new total length \( S' \) of the telescope when filled with water is:
\[ S' = f_o' + f_e' = \frac{R_o}{0.33} + \frac{R_e}{0.33} = \frac{R_o + R_e}{0.33} \]
Substituting \( R_o + R_e = 8 \, \text{cm} \):
\[ S' = \frac{8}{0.33} \approx 24.24 \, \text{cm} \]

### Step 8: Round to the nearest option
Since the options provided are 8 cm, 16 cm, 24 cm, and 32 cm, we round \( 24.24 \, \text{cm} \) to the nearest option, which is 24 cm.

### Final Answer
The present separation between the lenses when filled with water is **24 cm**.

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