In Young’s double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths $\lambda_1 = 12000 \AA$ and $\lambda_2 = 10000 \AA$. At what minimum distance from the common central bright fringe on the screen 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?
Interference was observed in interference chamber where air was present, now the chamber is evacuated, and if the same light is used, a careful observer will see
(1993)
When the chamber is evacuated, the refractive index decreases (from $\mu_{air}$ to 1). The wavelength of light $\lambda = \lambda_0/\mu$ increases. Since fringe width $\beta = \frac{\lambda D}{d}$, an increase in wavelength leads to a larger fringe width.
If yellow light emitted by sodium lamp in Young’s double slit experiment is replaced by monochromatic blue light of the same intensity
(1992)
Fringe width is given by $\beta = \frac{\lambda D}{d}$. Since the wavelength of blue light is less than that of yellow light ($\lambda_{blue} < \lambda_{yellow}$), the fringe width will decrease.
The angular resolution of a 10 cm diameter telescope at a wavelength of $5000 \AA$ is of the order of:
(2005)
The angular resolution limit is $\theta = \frac{1.22 \lambda}{D} = \frac{1.22 \times 5000 \times 10^{-10}}{10 \times 10^{-2}} = \frac{1.22 \times 5 \times 10^{-7}}{10^{-1}} = 6.1 \times 10^{-6} rad$. This is of the order of $10^{-6} rad$.
Diameter of human eye lens is 2 mm. What will be the minimum distance between two points to resolve them, which are situated at a distance of 50 metre from eye. The wavelength of light is $5000 \AA$:
(2002)
Using the simplified limit of resolution $\theta = \frac{\lambda}{D}$ (ignoring 1.22 factor to match options), we have $\frac{x}{d} = \frac{\lambda}{D}$, where $x$ is the minimum distance. $x = \frac{5000 \times 10^{-10}}{2 \times 10^{-3}} \times 50 = 2.5 \times 10^{-4} \times 50 = 1.25 \times 10^{-2} m = 1.25 cm$.
The Brewsters angle $i_b$ for an interface should be :
(2020)
According to Brewster's law, $\mu = \tan i_b$. For any typical optical interface from air to a denser medium, the refractive index $\mu > 1$. Therefore, $\tan i_b > 1$, which implies that $45^\circ < i_b < 90^\circ$.
Unpolarised light is incident from air on a plane surface of a material of refractive index ‘$\mu$’. At a particular angle of incidence ‘$i$’, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?
(2018)
When reflected and refracted rays are perpendicular, the light is incident at Brewster's angle, $i = \tan^{-1}(\mu)$. At this angle, the reflected light is completely plane-polarized with its electric field vector perpendicular to the plane of incidence.
Two Polaroids $P_1$ and $P_2$ are placed with their axis perpendicular to each other. Unpolarised light $I_0$ is incident on $P_1$. A third polaroid $P_3$ is kept in between $P_1$ and $P_2$ such that its axis makes an angle $45^\circ$ with that of $P_1$. The intensity of transmitted light through $P_2$ is:
(2017-Delhi)
Intensity after $P_1$ is $I_1 = \frac{I_0}{2}$. $P_3$ is at $45^\circ$ to $P_1$, so intensity after $P_3$ is $I_3 = I_1 \cos^2(45^\circ) = (\frac{I_0}{2})(\frac{1}{2}) = \frac{I_0}{4}$. $P_2$ is perpendicular to $P_1$, so it is at $45^\circ$ to $P_3$. Intensity after $P_2$ is $I_2 = I_3 \cos^2(45^\circ) = (\frac{I_0}{4})(\frac{1}{2}) = \frac{I_0}{8}$.
Which of the phenomenon is not common to sound and light waves?
(1988)
Sound waves are longitudinal mechanical waves, whereas light waves are transverse electromagnetic waves. Polarization is a property unique to transverse waves; therefore, longitudinal waves like sound cannot be polarized.
An electromagnetic radiation of frequency $n$, wavelength $\lambda$, travelling with velocity $v$ in air, enters a glass slab of refractive index $\mu$. The frequency, wavelength and velocity of light in the glass slab will be respectively
(1997)
When light passes from one medium to another, its frequency $n$ remains constant as it depends on the source. The velocity of light decreases to $v' = \frac{v}{\mu}$, and correspondingly, the wavelength decreases to $\lambda' = \frac{\lambda}{\mu}$.