
The incidence of a monochromatic light wave on a glass slab with a refractive index that increases linearly from \(n_1\) to \(n_2\) over a height \(h\) requires understanding how the light will behave as it exits the slab. When the refractive index varies, the exit angle of the light wave from the slab changes based on this variation. Let's analyze the situation:
1. Overall Concept: When the refractive index of a medium changes, the speed of light within that medium varies, resulting in refraction. For a linearly varying refractive index, this can cause the light to bend gradually.
2. Deflection Calculation: The light wave deflection angle, due to the refractive index gradient across the height of the slab, can be calculated using the expression for a linear refractive index gradient: \[ \theta = \tan^{-1}\left(\frac{(n_2-n_1)d}{h}\right) \]
Here, \(\theta\) is the angle of deflection, \(d\) is the thickness of the slab, and \(h\) is the height over which the refractive index changes from \(n_1\) to \(n_2\).
3. Key Observations:
Based on these calculations and observations, the correct statements are:
| It will deflect up by an angle \( \tan^{-1}\left(\frac{(n_2-n_1)d}{h}\right) \). |
| The deflection angle depends only on \( (n_2-n_1) \) and not on the individual values of \( n_1 \) and \( n_2 \). |
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Electromagnetic waves carry energy but not momentum.
Reason (R): Mass of a photon is zero. In the light of the above statements.
choose the most appropriate answer from the options given below:
The dimension of $ \sqrt{\frac{\mu_0}{\epsilon_0}} $ is equal to that of: (Where $ \mu_0 $ is the vacuum permeability and $ \epsilon_0 $ is the vacuum permittivity)
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
The waves that are produced when an electric field comes into contact with a magnetic field are known as Electromagnetic Waves or EM waves. The constitution of an oscillating magnetic field and electric fields gives rise to electromagnetic waves.
Electromagnetic waves can be grouped according to the direction of disturbance in them and according to the range of their frequency. Recall that a wave transfers energy from one point to another point in space. That means there are two things going on: the disturbance that defines a wave, and the propagation of wave. In this context the waves are grouped into the following two categories: