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.}
Step 1: Using the Refraction Formula for a Spherical Surface
The refraction formula for a spherical surface is given by: \[ \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \] where:
- \( n_1 \) is the refractive index of the first medium (air, \( n_1 = 1 \)),
- \( n_2 \) is the refractive index of the second medium (glass, \( n_2 = 1.5 \)),
- \( u \) is the object distance,
- \( v \) is the image distance,
- \( R \) is the radius of curvature.
Step 2: Substituting the Given Values
Since the object is at \( u = 0 \), the equation simplifies to: \[ \frac{n_2}{v} - \frac{1}{0} = \frac{1.5 - 1}{R} \] Since \( \frac{1}{0} \) tends to infinity, the equation reduces to: \[ \frac{n_2}{v} = \frac{0.5}{R} \]
Step 3: Solving for \( v \)
Rearranging the equation: \[ v = \frac{1.5 R}{0.5} = 3R \] Thus, the image distance is \( 3R \).
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 ……..
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 …….
For a plane mirror, the focal length is ……..
If the value of \( \cos \alpha \) is \( \frac{\sqrt{3}}{2} \), then \( A + A = I \), where \[ A = \begin{bmatrix} \sin\alpha & -\cos\alpha \\ \cos\alpha & \sin\alpha \end{bmatrix}. \]