A double convex thin lens made of glass (refractive index `mu = 1.5`) has both radii of curvature of magnitude 20 cm . Incident light rays parallel to the axis of the lens will converge at a distance L such that

Options

A. L=20 cm
B. L=10 cm
C. L = 40 cm
D. L = `20//3` cm

Answer

To find the distance \( L \) at which the incident light rays converge after passing through a double convex thin lens, we can use the lens maker's formula. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Refractive index of the lens \( \mu = 1.5 \)
- Radius of curvature \( R_1 = +20 \, \text{cm} \) (for the first surface)
- Radius of curvature \( R_2 = -20 \, \text{cm} \) (for the second surface, negative because it is opposite to the direction of light travel)

### Step 2: Use the lens maker's formula
The lens maker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Where:
- \( f \) is the focal length of the lens.

### Step 3: Substitute the values into the formula
Substituting the known values into the formula:
\[ \frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-20} \right) \]

### Step 4: Simplify the equation
Calculating \( \mu - 1 \):
\[ \mu - 1 = 0.5 \]
Now, calculating \( \frac{1}{20} - \frac{1}{-20} \):
\[ \frac{1}{20} + \frac{1}{20} = \frac{2}{20} = \frac{1}{10} \]
Now substituting back into the equation:
\[ \frac{1}{f} = 0.5 \times \frac{1}{10} = \frac{0.5}{10} = \frac{1}{20} \]

### Step 5: Calculate the focal length
Taking the reciprocal to find \( f \):
\[ f = 20 \, \text{cm} \]

### Step 6: Conclusion
The distance \( L \) at which the incident light rays converge is equal to the focal length \( f \):
\[ L = 20 \, \text{cm} \]

### Final Answer
Thus, the value of \( L \) is \( 20 \, \text{cm} \).

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