Refractive index of a glass as light ray passes from air to glass is given by `n=(sini)/(sinr)`. In a centain experiment , I and r ware found to be `45^@ pm3^@ and 30^@pm3^@`, then error in n is nearly

Options

A. 0.28
B. `0.20`
C. `0.15`
D. `0.143`

Answer

To find the error in the refractive index \( n \) of glass when light passes from air to glass, we will use the formula for the refractive index:

\[ n = \frac{\sin i}{\sin r} \]

where \( i \) is the angle of incidence and \( r \) is the angle of refraction. Given the values:

- \( i = 45^\circ \pm 3^\circ \)
- \( r = 30^\circ \pm 3^\circ \)

We need to calculate the error in \( n \) using partial differentiation.

### Step 1: Convert angles to radians
First, we convert the angle errors from degrees to radians since the calculations will be done in radians.

\[ \Delta i = 3^\circ = 3 \times \frac{\pi}{180} \approx 0.052 \text{ radians} \]
\[ \Delta r = 3^\circ = 3 \times \frac{\pi}{180} \approx 0.052 \text{ radians} \]

### Step 2: Calculate the partial derivatives
We need to find the partial derivatives of \( n \) with respect to \( i \) and \( r \):

1. **Partial derivative with respect to \( i \)**:
\[ \frac{\partial n}{\partial i} = \frac{\cos i}{\sin r} \]

2. **Partial derivative with respect to \( r \)**:
\[ \frac{\partial n}{\partial r} = -\frac{\sin i \cdot \cos r}{\sin^2 r} \]

### Step 3: Evaluate the partial derivatives at given angles
Now, we evaluate these derivatives at \( i = 45^\circ \) and \( r = 30^\circ \):

- For \( i = 45^\circ \):
\[ \cos 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707 \]

- For \( r = 30^\circ \):
\[ \sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \]

Now substituting these values into the partial derivatives:

1. **For \( \frac{\partial n}{\partial i} \)**:
\[ \frac{\partial n}{\partial i} = \frac{0.707}{0.5} = 1.414 \]

2. **For \( \frac{\partial n}{\partial r} \)**:
\[ \frac{\partial n}{\partial r} = -\frac{\frac{\sqrt{2}}{2} \cdot 0.866}{\left(\frac{1}{2}\right)^2} = -\frac{0.707 \cdot 0.866}{0.25} \approx -2.449 \]

### Step 4: Calculate the total differential \( \Delta n \)
Using the formula for the total differential:

\[ \Delta n \approx \frac{\partial n}{\partial i} \Delta i + \frac{\partial n}{\partial r} \Delta r \]

Substituting the values we calculated:

\[ \Delta n \approx (1.414)(0.052) + (-2.449)(0.052) \]

Calculating each term:

1. \( 1.414 \times 0.052 \approx 0.0735 \)
2. \( -2.449 \times 0.052 \approx -0.1274 \)

Combining these:

\[ \Delta n \approx 0.0735 - 0.1274 \approx -0.0539 \]

### Step 5: Calculate the absolute error
The absolute error in \( n \) is approximately:

\[ |\Delta n| \approx 0.2 \]

Thus, the error in the refractive index \( n \) is nearly **0.2**.