The refractive index of a particular material is 1.67 for blue light, 1.65 for yellow light and 1.63 for red light. The dispersive power of the material is .........

Options

A. (a)0.0615
B. (b)0.024
C. (c)0.031
D. (d)1.6

Answer

To find the dispersive power of the material, we can use the formula for dispersive power (ω):

\[ \omega = \frac{\mu_v - \mu_r}{\mu_y - 1} \]

where:
- \(\mu_v\) is the refractive index for blue light,
- \(\mu_r\) is the refractive index for red light,
- \(\mu_y\) is the refractive index for yellow light.

### Step 1: Identify the refractive indices
From the problem, we have:
- \(\mu_b = 1.67\) (for blue light),
- \(\mu_y = 1.65\) (for yellow light),
- \(\mu_r = 1.63\) (for red light).

### Step 2: Substitute the values into the formula
Now, we can substitute these values into the dispersive power formula:

\[ \omega = \frac{1.67 - 1.63}{1.65 - 1} \]

### Step 3: Calculate the numerator
Calculate the numerator:

\[ 1.67 - 1.63 = 0.04 \]

### Step 4: Calculate the denominator
Calculate the denominator:

\[ 1.65 - 1 = 0.65 \]

### Step 5: Calculate dispersive power
Now, substitute the values back into the formula:

\[ \omega = \frac{0.04}{0.65} \]

### Step 6: Perform the division
Now, perform the division:

\[ \omega = 0.061538 \approx 0.0615 \]

### Conclusion
Thus, the dispersive power of the material is approximately \(0.0615\).