The two surfaces of a biconvex lens has same radii of curvatures . This lens is made of glass of refractive index 1.5 and has a focal length of 10 cm in air. The lens is cut into two equal halves along a plane perpendicular to its principal axis to yield two plane - convex lenses. The two pieces are glued such that the convex surfaces touch each other. If this combination lens is immersed in water (refractive index =`4/3` ), its focal length (in cm ) is

Options

A. 5 cm
B. 10 cm
C. 20 cm
D. 40 cm

Answer

To solve the problem, we need to determine the focal length of the combination lens when immersed in water. Here are the steps to find the solution:

### Step 1: Understand the Initial Configuration
The original biconvex lens has a focal length \( f = 10 \, \text{cm} \) in air. The lens is made of glass with a refractive index \( n_g = 1.5 \).

### Step 2: Calculate the Radius of Curvature
Using the lens maker's formula for a biconvex lens:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Since both radii of curvature are equal (\( R_1 = R_2 = R \)), we can rewrite the formula as:
\[ \frac{1}{f} = (n_g - 1) \left( \frac{2}{R} \right) \]
Substituting \( n_g = 1.5 \) and \( f = 10 \, \text{cm} \):
\[ \frac{1}{10} = (1.5 - 1) \left( \frac{2}{R} \right) \]
\[ \frac{1}{10} = 0.5 \left( \frac{2}{R} \right) \]
\[ \frac{1}{10} = \frac{1}{R} \]
Thus, \( R = 10 \, \text{cm} \).

### Step 3: Determine the Focal Length of the Plane-Convex Lens
When the biconvex lens is cut into two equal halves, each half becomes a plane-convex lens. The focal length of a plane-convex lens is given by:
\[ f' = \frac{R}{2(n - 1)} \]
For the plane-convex lens made of glass:
\[ f' = \frac{10}{2(1.5 - 1)} = \frac{10}{2 \times 0.5} = \frac{10}{1} = 10 \, \text{cm} \]

### Step 4: Combine the Two Plane-Convex Lenses
When the two plane-convex lenses are glued together with their convex surfaces touching, they act as a single lens. The focal length \( f_{comb} \) of two lenses in contact is given by:
\[ \frac{1}{f_{comb}} = \frac{1}{f'} + \frac{1}{f'} \]
\[ \frac{1}{f_{comb}} = \frac{2}{10} = \frac{1}{5} \]
Thus, the focal length of the combination lens in air is:
\[ f_{comb} = 5 \, \text{cm} \]

### Step 5: Determine the Focal Length in Water
Now, we need to find the focal length of the combination lens when it is immersed in water. The effective refractive index \( n_{eff} \) of the lens in water is given by:
\[ n_{eff} = \frac{n_g}{n_w} = \frac{1.5}{\frac{4}{3}} = \frac{1.5 \times 3}{4} = \frac{4.5}{4} = 1.125 \]
Using the lens maker's formula again:
\[ \frac{1}{f_{water}} = (n_{eff} - 1) \left( \frac{2}{R} \right) \]
Substituting \( R = 10 \, \text{cm} \):
\[ \frac{1}{f_{water}} = (1.125 - 1) \left( \frac{2}{10} \right) \]
\[ \frac{1}{f_{water}} = 0.125 \left( \frac{2}{10} \right) = 0.125 \times 0.2 = 0.025 \]
Thus, the focal length in water is:
\[ f_{water} = \frac{1}{0.025} = 40 \, \text{cm} \]

### Final Answer
The focal length of the combination lens when immersed in water is \( 40 \, \text{cm} \).

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