A double convex lens is made of glass which has refractive inded 1.55 for voilet rays and 1.50 for red rays. If the focal length for violet rays is 20 cm, the focal length for red rays will be

Options

A. 9 cm
B. 28 cm
C. 20 cm
D. 22 cm

Answer

To find the focal length of the red rays using the given information, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Refractive index for violet rays (\( \mu_v \)) = 1.55
- Refractive index for red rays (\( \mu_r \)) = 1.50
- Focal length for violet rays (\( f_v \)) = 20 cm

2. **Use the Lens Maker's Formula:**
The lens maker's formula is given by:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Here, \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces. Since the lens is the same for both violet and red rays, \( R_1 \) and \( R_2 \) will be the same for both cases.

3. **Set Up the Equations:**
For violet rays:
\[ \frac{1}{f_v} = (\mu_v - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \tag{1} \]
For red rays:
\[ \frac{1}{f_r} = (\mu_r - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \tag{2} \]

4. **Divide Equation (1) by Equation (2):**
\[ \frac{f_r}{f_v} = \frac{\mu_v - 1}{\mu_r - 1} \]

5. **Rearrange to Find \( f_r \):**
\[ f_r = f_v \cdot \frac{\mu_v - 1}{\mu_r - 1} \]

6. **Substitute the Known Values:**
- \( f_v = 20 \) cm
- \( \mu_v - 1 = 1.55 - 1 = 0.55 \)
- \( \mu_r - 1 = 1.50 - 1 = 0.50 \)

Now substituting these values:
\[ f_r = 20 \cdot \frac{0.55}{0.50} \]

7. **Calculate \( f_r \):**
\[ f_r = 20 \cdot 1.1 = 22 \text{ cm} \]

### Final Answer:
The focal length for red rays (\( f_r \)) is **22 cm**.

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