A double convex lens of focal length 20 cm is made of glass of refractive index 3 / 2. When placed completely in water `(._(a)mu_(w)=4//3)`, its focal length will be

Options

A. 80 cm
B. 15 cm
C. 17.7 cm
D. 22.5 cm

Answer

To solve the problem of finding the new focal length of a double convex lens when placed in water, we can use the lens maker's formula. Here is the step-by-step solution:

### Step 1: Understand the Lens Maker's Formula
The lens maker's formula is given by:

\[ \frac{1}{f} = \left( \mu - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Where:
- \( f \) is the focal length of the lens,
- \( \mu \) is the refractive index of the lens material relative to the medium,
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 2: Calculate the Refractive Index of Glass with Respect to Air
Given:
- Focal length of the lens in air, \( f = 20 \) cm,
- Refractive index of glass, \( \mu_g = \frac{3}{2} = 1.5 \).

Using the lens maker's formula in air:

\[ \frac{1}{20} = \left( \frac{3}{2} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

This simplifies to:

\[ \frac{1}{20} = \left( \frac{1}{2} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

### Step 3: Rearranging the Equation
From the equation, we can express \( \frac{1}{R_1} - \frac{1}{R_2} \):

\[ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{20} \times 2 = \frac{1}{10} \]

### Step 4: Calculate the Refractive Index of Glass with Respect to Water
Now, when the lens is placed in water, the effective refractive index \( \mu_{gw} \) is given by:

\[ \mu_{gw} = \frac{\mu_g}{\mu_w} = \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8} = 1.125 \]

### Step 5: Apply the Lens Maker's Formula in Water
Now we apply the lens maker's formula again, but this time for the lens in water:

\[ \frac{1}{f_w} = \left( \mu_{gw} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

Substituting the values:

\[ \frac{1}{f_w} = \left( \frac{9}{8} - 1 \right) \left( \frac{1}{10} \right) \]

This simplifies to:

\[ \frac{1}{f_w} = \left( \frac{1}{8} \right) \left( \frac{1}{10} \right) = \frac{1}{80} \]

### Step 6: Calculate the New Focal Length
Taking the reciprocal gives us the new focal length:

\[ f_w = 80 \text{ cm} \]

### Final Answer
The focal length of the lens when placed completely in water is **80 cm**.

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