For a thin prism, if the angle of the prism is \( A \) with a refractive index of 1.6, then the angle of minimum deviation will be …….
Step 1: Understanding the Minimum Deviation Formula
For a thin prism, the formula for the angle of minimum deviation (\( \delta_m \)) is: \[ \delta_m = (n - 1) A \] where:
- \( n \) is the refractive index of the prism,
- \( A \) is the angle of the prism,
- \( \delta_m \) is the minimum deviation.
Step 2: Substituting the Given Values
Given: - \( n = 1.6 \), - \( A = 4^\circ \).
Step 3: Calculating Minimum Deviation
\[ \delta_m = (1.6 - 1) \times 4^\circ \] \[ \delta_m = 0.6 \times 4^\circ \] \[ \delta_m = 2.4^\circ \] Thus, the angle of minimum deviation is \( 2.4^\circ \).
The refractive index of glass is 1.6 and the speed of light in glass will be ……… . The speed of light in vacuum is \( 3.0 \times 10^8 \) ms\(^{-1}\).
Consider a refracting telescope whose objective has a focal length of 1m and the eyepiece a focal length of 1cm, then the magnifying power of this telescope will be ……..
A ray coming from an object which is situated at zero distance in the air and falls on a spherical glass surface (\( n = 1.5 \)). Then the distance of the image will be ………. \( R \) is the radius of curvature of a spherical glass.}
For a plane mirror, the focal length is ……..
If the value of \( \cos \alpha \) is \( \frac{\sqrt{3}}{2} \), then \( A + A = I \), where \[ A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}. \]