To solve this problem, we need to recall the formula used for refraction at a spherical surface separating two media of different refractive indices.
1. Refraction at Spherical Surface:
When a light ray travels from a medium of refractive index \( n_1 \) to another medium with refractive index \( n_2 \), and the interface is a curved surface of radius \( R \), the refraction formula is:
\( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \)
2. Meaning of Symbols:
- \( u \): object distance
- \( v \): image distance
- \( R \): radius of curvature
- \( n_1 \): refractive index of medium where object is located
- \( n_2 \): refractive index of the second medium
3. Evaluating the Options:
(1) \( \frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R} \) – Correct ✅
(2) \( v - u = \frac{n_2 - n_1}{R} \) – Incorrect (dimensionally inconsistent)
(3) \( \frac{v}{n_2} - \frac{u}{n_1} = \frac{n_2 - n_1}{R} \) – Incorrect
(4) \( \frac{n_2}{v} + \frac{n_1}{u} = \frac{n_2 - n_1}{R} \) – Incorrect (wrong sign)
Final Answer:
The correct option is (A).
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = 4/3 \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is cm. 