Question

The radii of curvatures of the focus of a double convex lens are 10 cm and 15 cm and its focal length is 12 cm. What is the refractive index of glass? (Refractive index of air = 1)
OR
A convex lens has 20 cm focal length in air. What is focal length in water? (Refractive index of air-water = 1.33, refractive index for air-glass = 1.5)

Hint:

The focal length of a lens decreases when it is placed in a medium with a higher refractive index compared to air.

Answer 1

a. Refractive Index of Glass:
We can use the lens maker's formula to calculate the refractive index of glass. The lens maker's formula is:
\[ \frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Where:
- \( f \) is the focal length of the lens,
- \( n \) is the refractive index of the lens material (glass in this case),
- \( R_1 \) and \( R_2 \) are the radii of curvature of the two surfaces of the lens.
From the problem, we are given:
- \( R_1 = 10 \, \text{cm} \),
- \( R_2 = -15 \, \text{cm} \) (since the second radius is on the opposite side of the first),
- \( f = 12 \, \text{cm} \),
- Refractive index of air is \( 1 \).
Substitute the given values into the lens maker's formula:
\[ \frac{1}{12} = (n - 1) \left( \frac{1}{10} - \frac{1}{-15} \right) \] \[ \frac{1}{12} = (n - 1) \left( \frac{1}{10} + \frac{1}{15} \right) \] \[ \frac{1}{12} = (n - 1) \left( \frac{5}{30} + \frac{2}{30} \right) \] \[ \frac{1}{12} = (n - 1) \times \frac{7}{30} \] \[ (n - 1) = \frac{30}{12 \times 7} = \frac{30}{84} = \frac{5}{14} \] \[ n = 1 + \frac{5}{14} = \frac{19}{14} \] \[ n = 1.357 \]
Thus, the refractive index of glass is approximately \( 1.357 \).
b. Focal Length in Water:
The focal length of a lens in a medium other than air is related to its focal length in air by the following formula:
\[ \frac{f_{\text{medium}}}{f_{\text{air}}} = \frac{n_{\text{air}}}{n_{\text{medium}}} \]
For this problem:
- \( f_{\text{air}} = 20 \, \text{cm} \),
- The refractive index of air is \( n_{\text{air}} = 1 \),
- The refractive index of water is \( n_{\text{water}} = 1.33 \).
We can rearrange the equation to find the focal length in water:
\[ f_{\text{water}} = f_{\text{air}} \times \frac{n_{\text{air}}}{n_{\text{water}}} \]
Substitute the given values:
\[ f_{\text{water}} = 20 \times \frac{1}{1.33} \approx 15.04 \, \text{cm} \]
Thus, the focal length of the lens in water is approximately \( 15.04 \, \text{cm} \).