Question
Unpolarized light is incident on two polarizers. The axes of polarization of both polarizers are initially parallel.
The second polarizer is then rotated through 360° as shown.
Answer – B
According to Law of Malus : \(I α cos^{2}θ\), where I is intensity of light after second polariser
According to this equation emergent intensity will be maximum corresponding to angles \(0^{\circ}\) and \(180^{\circ}\) and minimu for \(90^{\circ} \)and \(270^{\circ}\). So option B is correct
Question
Unpolarized light is incident on two polarizing filters X and Y. They are arranged so that light emerging from Y has a maximum intensity. X is fixed and Y is rotated through about the direction of the incident beam in its own plane.
Answer – C
According to Law of Malus : \(I α cos^{2} θ\), where I is intensity of light after second polariser
Cosθ will be max at \(180^{\circ}\) ,\(360^{\circ}\) ,\(540^{\circ}\)
Question
Unpolarized light of intensity lois incident on a polarizing filter. Light from this filter is incident on a second filter, which has its axis of polarization at 30° to that of the first filter.
Answer – D
The unpolarized light of Intensity \(I_{0}\)
is incident on a polarizer then a polarized light at intensity \(I_{0}/2\) comes out.
Intensity of transmitted beam
\(I = I_{0}/2 cos ^{2} 30°\)
\(I= 3 I_{0}/8\)
Question
Which diagram shows the shape of the wavefront as a result of the diffraction of plane waves by an object?
Answer – A
Diffraction is the spreading out of waves as they pass through an aperture or around objects. It occurs when the size of the aperture or obstacle is of the same order of magnitude as the wavelength of the incident wave. For very small aperture sizes, the vast majority of the wave is blocked.
Spherical,
The point source emits the spherical wavefront as the light diverges from a point source in all directions.
Question
A point source of light of amplitude Ao gives rise to a particular light intensity when viewed at a distance from the source. When the amplitude is increased and the viewing distance is doubled, the light intensity is doubled. What is the new amplitude of the source?
Answer – B
From the two proportional rules
\(I=k\frac{A^{2}}{X^{2}}\) where k is constant
\(2I=k\frac{{A}’^{2}}{(2X)^{2}}\)
isolating A and {A}’ and dividing we obtain
\(\frac{{A}’}{A}=2\sqrt{2}\)
Question
Horizontally polarized light is transmitted through a polarizer whose transmission axis is horizontal. The light enters a container with a sugar solution and is then incident on a second polarizer whose
transmission axis is vertical.
Answer – B
Question
Unpolarized light is shone through two identical polarizers whose axes are parallel.
Answer – B
Since the incident light is unpolarized, the intensity of the incident light halves when it passes through the first polarizer: I/2. This is a consequence of the average value of the cosine squared function (not necessary to know). As the second polarizer is parallel to the first, the intensity remains unchanged when it passes through the second polarizer. I.e., the transmitted intensity is \(I/2\). So:
\(0.5I/I = 0.5\), which is equivalent to 50%
Question
Wave generators placed at position P and position Q produce water waves of wavelength 4.0 cm. Each generator, operating alone, produces a wave oscillating with amplitude A at position R. Distances PR and QR are shown in the diagram below.
Answer – A
\(\Lambda =4 cm\)
\(QR= 14 cm= 3\Lambda +\frac{\Lambda }{2}\)
\(PR=20 cm= 5\Lambda \)
Path diffrence \(= 5\Lambda -\frac{7\Lambda }{2}\)
Phase diff. \(= \frac{2\Pi }{\Lambda }\times \frac{3\Lambda }{2}\rightarrow 3\Pi \)
and \()Cos3\Pi =0\)
so \(R^{2}=A^{2}+A^{2}+2A.Acos3\Pi \rightarrow R=0\)
Question
An unpolarized ray of light in air is incident on the surface of water. The reflected ray is completely polarized. Which of the following are separated by an angle of 90°?
Answer – B
An unpolarized light beam is incident on a surface at an angle of incidence equal to Brewster’s angle. Then, the reflected beam used to polarized completely and the refracted beam used to be polarized partially. Also, both these beams used to be at right angle to each other.
Question
Unpolarized light of intensity lois incident on a polarizing filter. Light from this filter is incident on a second filter, which has its axis of polarization at 30° to that of the first filter.
Answer – D
The unpolarized light of Intensity \(I_{0}\) is incident on a polarizer then a polarized light at intensity I_{0}/2 comes out.
Intensity of transmitted beam
\(I = I_{0}/2 cos ^{2} 30°\)
\(I= 3 I_{0}/8\)
Question
Unpolarized light of intensity Io is incident on a polarizer that has a vertical transmission axis.
Answer – D
When unpolarized light is transmitted through a polaroid filter, it emerges with one-half the intensity(constant)and with vibrations in a single plane, it emerges as a polarized light.
The unpolarized light of Intensity \(I_{0}\)
is incident on a polarizer then a polarized light at intensity \(I_{0}/2\) comes out.
Intensity of transmitted beam
Question
Monochromatic coherent light is incident on a narrow rectangular slit. The diffracted light is observed on a distant screen. The graph below shows how the intensity of the light varies with position on the screen.
Answer – D
Question
A person wearing polarizing sunglasses stands at the edge of a pond in bright sunlight.
Answer – D
Brewster’s angle (also known as the polarization angle) is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection.
Brewster’s angle \(= tan^{-1} (n2/n1)\)
where \(n_{1}\) is the refractive index of the initial medium through which the light propagates (the “incident medium”), and \(n_{2}\) is the index of the other medium.
So here value on n1 = 1
And n2 = n
\(tan\Theta =\frac{1}{n}\)
Question
Polarized light of intensity/o is incident on a polarizing filter. The angle between the plane of polarization of the incident light and the transmission plane of the polarizer is 9. Which graph shows how the intensity of the light transmitted through the polarizer varies with e?
Answer – D
Question
Two polarizing filters are set up so the transmitted light is at a maximum intensity.
Answer – C
According to Law of Malus :\( I\alpha \cos ^{2}\Theta \) , where I is intensity of light after second polariser
For no light \( \cos \Theta =0\) which is at \(90^{\circ}\)
Question
Unpolarized light of intensity lo is transmitted through a polarizer which has a transmission axis at an angle to the vertical. The light is then incident on a second polarizer with a transmission axis at an angle to the transmission axis of the first polarizer, as shown below.
Answer – C
According to Law of Malus :\( I\alpha \cos ^{2}\Theta \), where I is intensity of light after second polariser.
The unpolarized light of Intensity \(I_{O} \) is incident on a polarizer then a polarized light at intensity \(\frac{I_{o}}{2}\) comes out.
Intensity of transmitted beam
Question
Unpolarized light is incident on a polarizer. The light transmitted by the first polarizer is then incident on a second polarizer. The polarizing axis of the second polarizer is at 60° to that of the first polarizer.
Answer – D
Let
The unpolarized light of Intensity I_{0} is incident on a polarizer then a polarized light at intensity I_{0}/2 comes out.
Intensity of transmitted beam
\(I_{f}=\frac{I_{o}}{2}\times \cos ^{2}60^{\circ}\)
\(I_{f}=\frac{I_{o}}{8}\)
\(I_{o}=8I_{f}\)
Question
Unpolarized light of intensity lo is incident on a polarizer with a vertical transmission axis. The transmitted light is incident on a sheet of material X. After transmission through X the intensity of the light is
Answer -A
Question
Plane-polarized light is incident normally on a polarizer which is able to rotate in the plane perpendicular to the light as shown below.
Answer – C
Question
Plane-polarized light is incident normally on a polarizer which is able to rotate in the plane perpendicular to the light as shown below.
Answer – C
\(I=I_{o}\cos ^{2}\Theta\)
\(2=8\cos ^{2}\Theta\)
\(\sqrt{0.25}=\sqrt{\cos ^{2}\Theta }\)
\(\Theta =60^{\circ}\)
Now it. Rotated by 90°
\(90+60=150^{\circ}\)
\(I=I_{o}\cos ^{2}\Theta \)
\(I=8\cos ^{2}150^{\circ}\)
\(I=6Wm^{-2}\)