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 $\lambda' = \frac{\lambda}{\mu}$.
A beam of light of $lambda = 600 nm$ from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is:
(2014)
The distance between the first dark fringes on either side of the central bright fringe is the linear width of the central maximum, given by $W = \frac{2D\lambda}{a}$. Substituting the values, $W = \frac{2 \times 2 \times 600 \times 10^{-9}}{1 \times 10^{-3}} = 2400 \times 10^{-6} m = 2.4 mm$.
A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct?
(2013)
The de Broglie wavelength of an electron is $\lambda = \frac{h}{mv}$. As the speed $v$ increases, the wavelength $\lambda$ decreases. The angular width of the central maximum is $2\theta = \frac{2\lambda}{a}$. Since $\lambda$ decreases, the angular width will decrease.
A parallel beam of monochromatic light of wavelength $5000 \AA$ is incident normally on a single narrow slit of width 0.001 mm. The light is focused by a convex lens on a screen placed in focal plane. The first minimum will be formed for the angle of diffraction equal to
(1993)
For the first minimum in a single slit diffraction pattern, $a \sin\theta = \lambda$. Thus, $\sin\theta = \frac{\lambda}{a} = \frac{5000 \times 10^{-10}}{0.001 \times 10^{-3}} = \frac{5 \times 10^{-7}}{10^{-6}} = 0.5$. Therefore, $\theta = \sin^{-1}(0.5) = 30^\circ$.
Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is:
(2020)
The limit of resolution of a telescope is given by $\Delta\theta = \frac{1.22 \lambda}{D}$. Substituting the values, $\Delta\theta = \frac{1.22 \times 600 \times 10^{-9}}{2} = 1.22 \times 300 \times 10^{-9} = 3.66 \times 10^{-7} rad$.
An astronomical refracting telescope will have large angular magnification and high angular resolution, when it has an objective lens of
(2018)
Angular magnification is $M = \frac{f_o}{f_e}$, requiring a large focal length $f_o$ for the objective. Angular resolution is inversely proportional to the resolving limit $\Delta\theta = \frac{1.22\lambda}{D}$, requiring a large aperture diameter $D$ for high resolution.