The focal length of a converting lens are `f_(v) and f_(r)` for violet and red lights, respectively. Which of the following is correct?

Options

A. `f_(v)ltF_(r)`
B. `f_(v)=F_(r)`
C. `f_(v)gtF_(r)`
D. It depends on the average refractive index

Explanation

To solve the question regarding the focal lengths \( f_v \) and \( f_r \) of a converging lens for violet and red lights respectively, we can follow these steps:

### Step 1: Understand the Relationship Between Focal Length and Refractive Index
The focal length \( f \) of a lens is related to the refractive index \( \mu \) of the material of the lens and the radii of curvature \( R_1 \) and \( R_2 \) of the lens surfaces by the lens maker's formula:
\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Here, \( \mu \) is the refractive index of the lens material for the specific wavelength of light.

### Step 2: Compare the Refractive Indices of Violet and Red Light
The refractive index of a material generally increases as the wavelength of light decreases. Therefore, for violet light (which has a shorter wavelength) the refractive index \( \mu_v \) is greater than that for red light \( \mu_r \):
\[ \mu_v > \mu_r \]

### Step 3: Analyze the Effect on Focal Length
From the lens maker's formula, since \( \mu_v > \mu_r \), it follows that:
\[ \frac{1}{f_v} = (\mu_v - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
\[ \frac{1}{f_r} = (\mu_r - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]
Since \( \mu_v > \mu_r \), it implies that \( f_v < f_r \). This means that the focal length for violet light is less than that for red light.

### Step 4: Conclusion
Thus, the correct relationship is:
\[ f_v < f_r \]

### Final Answer
The focal length of the converging lens for violet light \( f_v \) is less than that for red light \( f_r \).

---