As shown in the diagram, when the incident ray is parallel to base of the prism, the emergent ray grazes along the second surface. If refractive index of the material of prism is $\sqrt{2}$, the angle $\theta$ of prism is :

Question

$Li^{+2} \longrightarrow Li^{+3} + e^-$ :
Energy required for this process.
Given ionisation energy for ground state of hydrogen atom is $2.17 \times 10^{-18} \text{ J}$.

Answer 1

To solve this problem, we need to use the principles of geometric optics, specifically related to prisms and refraction.

The problem states that the emergent ray grazes along the second surface, meaning it makes an angle of 90° with the normal. The given refractive index of the prism material is \sqrt{2}.

Let the angle of the prism be \theta. According to Snell's Law:

n_1 \sin i = n_2 \sin r

Given that the refractive index n_1 (air) is 1 and n_2 (prism) is \sqrt{2}, and the incident angle i = 0 since the incident ray is along the normal:

1 \cdot \sin i = \sqrt{2} \cdot \sin 45^\circ

Thus, the angle of refraction r is 45° because:

\sin 45^\circ = \frac{1}{\sqrt{2}}

At the second surface, where the emergent angle e = 90^\circ, we apply the condition for grazing emergence:

\sin(90^\circ) = \frac{1}{\sqrt{2}}\sin(\theta)

Since \sin(90^\circ) = 1, we have:

1 = \frac{1}{\sqrt{2}}\sin(\theta)

Solving for \theta gives:

\sin(\theta) = \sqrt{2} \cdot 1

Therefore, \theta = 45^\circ.

Hence, the correct answer is 45°.

Answer 2