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  • An experiment is conducted to determine the refractive index of a glass prism using a spectrometer. A monochromat

An experiment is conducted to determine the refractive index(n) of a glass prism using a spectrometer. A monochromatic light of known wavelength (\lambda) is used. The prism is placed on a table, and the angle of minimum deviation (D) is measured. Calculate the refractive index of the glass prism.

Option: 1

5.154


Option: 2

3.126


Option: 3

6.255


Option: 4

7.667


Answers (1)

best_answer

Given values:
• Wavelength of light (\lambda)=589\ nm
• Angle of minimum deviation (D)=30^{\degree}

The refractive index (n) of the glass prism can be calculated using the for-
mula for the angle of minimum deviation:

n=\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{D}{2}\right)}

Where A is the angle of the prism.
For a prism, A+D=180{\degree}, so A=180{\degree}-D
Substitute the value of A into the formula for n

:n=\frac{\sin \left(\frac{180^{\circ}-D+D}{2}\right)}{\sin \left(\frac{D}{2}\right)}
Step 1: Calculate A:

A=180{\degree}-30{\degree}=150{\degree}
Step 2: Calculate the refractive index n:

\begin{gathered} n=\frac{\sin \left(\frac{150^{\circ}+30^{\circ}}{2}\right)}{\sin \left(\frac{30^{\circ}}{2}\right)} \\ n=\frac{\sin \left(\frac{180^{\circ}}{2}\right)}{\sin \left(\frac{15^{\circ}}{2}\right)} \\ n=\frac{1}{\sin \left(\frac{15^{\circ}}{2}\right)} \end{gathered}


Step 3: Calculate n:

\begin{gathered} n \approx \frac{1}{0.13053} \\ n \approx 7.667 \end{gathered}

The refractive index of the glass prism is approximately 7.667.
Therefore, the correct option is (D).

Posted by

sudhir.kumar

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