WAVE OPTICS Quiz-12
Dear Readers,
As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background.
Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.
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Q1. The wavelength of the matter waves is independent of
Solution
Wavelength of matter wave or de Broglie wave length 𝜆 = ℎ/𝑚𝑣 = ℎ/𝑝 From the above relation it is clear that wavelength of matter wave is independent the charge
Wavelength of matter wave or de Broglie wave length 𝜆 = ℎ/𝑚𝑣 = ℎ/𝑝 From the above relation it is clear that wavelength of matter wave is independent the charge
Q2. The graph showing the dependence of intensity of transmitted light on the angle between polarizer and analyser, is
Solution
According to law of Malus, when a beam of completely plane polarized light is incident on an analyser, the resultant intensity of light (𝐼) transmitted from the analyser varies directly as the square of the cosine of the angle (𝜃) between planes of transmission of analyser and polarizer 𝑖𝑒,𝐼 ∝ cos2 𝜃 and 𝐼 = 𝐼0 cos2 𝜃 …(i)
According to law of Malus, when a beam of completely plane polarized light is incident on an analyser, the resultant intensity of light (𝐼) transmitted from the analyser varies directly as the square of the cosine of the angle (𝜃) between planes of transmission of analyser and polarizer 𝑖𝑒,𝐼 ∝ cos2 𝜃 and 𝐼 = 𝐼0 cos2 𝜃 …(i)
Where 𝐼0 = intensity of the light from polarizer From Eq. (i), we note that if the transmission axes of polarizer and analyser are parallel (𝑖𝑒,𝜃 = 0° or 180°), then 𝐼 = 𝐼0. It means that intensity of transmitted light is maximum. When the transmission axes of polarizer and analyser are perpendicular (𝑖𝑒,𝜃 = 90°), then 𝐼 = 𝐼0 cos2 90° = 0. It means the intensity of transmitted light is minimum On plotting a graph between 𝐼 and 𝜃 as given by relation (i), we get the curve as shown in figure
Q3. In young’s double slit experiment 𝑑 𝐷
= 10-4(𝑑 =distance between slits, 𝐷 = distance of screen from the slits). At a point 𝑃 on the screen resultant intensity is equal to the intensity due to the individual slit𝐼0. Then the distance of point 𝑃 from the central maximum is (𝜆 = 6000 Å)
Solution
The resultant intensity at any point P is 𝐼 = 4𝐼0 cos2 ( ϕ/2 )
The resultant intensity at any point P is 𝐼 = 4𝐼0 cos2 ( ϕ/2 )
∴ 𝐼0 = 4𝐼0 cos2 ϕ/2
Or cos
ϕ/2
=
1/2
∴
ϕ/2
=
𝜋/3
𝑜𝑟 ϕ =
2𝜋/3
If Δ𝑥 is the corresponding value of path difference at P , then
ϕ =
2𝜋/𝜆
(Δ𝑥)
2𝜋/3
=
2𝜋/𝜆 Δx.
As Δ𝑥 =
𝑥𝑑/𝐷
∴
1/3
=
1/𝜆
𝑥𝑑/𝐷
Or 𝑥 =
𝜆 3𝑑/𝐷
=
6×10-7/3×10-4
= 2 × 10-3m
𝑥 = 2mm
This is the difference of point P from central maximum
Q4. . If white light is used in the Newton’s rings experiment, the colour observed in the reflected light is complementary to that observed in the transmitted light is complementary to that observed in the transmitted light through the same point. This is due to
Solution
The rings observed in reflected light are exactly complementary to those seen in transmitted light. Corresponding to every dark ring in reflected light there is a bright ring in transmitted light. The ray reflected at the upper surface of the air-film suffers no phase change while the ray reflected internally at the lower surface suffers a phase change of 𝜋.
The rings observed in reflected light are exactly complementary to those seen in transmitted light. Corresponding to every dark ring in reflected light there is a bright ring in transmitted light. The ray reflected at the upper surface of the air-film suffers no phase change while the ray reflected internally at the lower surface suffers a phase change of 𝜋.
Q5. In Fresnel’s biprism (𝜇 = 1.5) experiment the distance between source and biprism is 0.3 𝑚 and that between biprism and screen is 0.7𝑚 and angle of prism is 1°. The fringe width with light of wavelength 6000 Å will be
and 1/𝑣 is of the form
Solution
𝛽 = (𝑎 + 𝑏)𝜆 /2𝑎(𝜇 − 1)𝛼
Where 𝑎 = distance between source and biprism = 0.3 𝑚 𝑏 = distance between biprism and screen = 0.7 𝑚 𝛼 = Angle of prism = 1°,𝜇 = 1.5,𝜆 = 6000 × 10-10𝑚
Hence, 𝛽 =
(0.3+0.7)×6×10-7 /2×0.3(1.5−1)×(1°× 𝜋 180
)
= 1.14 × 10-4𝑚 = 0.0114 𝑐𝑚
.
Q6. Two light sources are said to be coherent if they are obtained from
Solution
When two sources are obtained from a single source, the wavefront is divided into two parts. These two wavefronts act as if they are emanated from two sources having a fixed phase relationship
When two sources are obtained from a single source, the wavefront is divided into two parts. These two wavefronts act as if they are emanated from two sources having a fixed phase relationship
Q7. When the angle of incidence on a material is 60°, the reflected light is completely polarized. The velocity of the refracted ray inside the material is (in ms-1)
Solution
From Brewster’s law,
From Brewster’s law,
𝜇 = tan𝑖𝑝
⇒
𝑐/𝑣
= tan60° = √3
⇒ 𝑣 =𝑐/√3
=
3 × 108 /√3
= √3 ×
108m/s
Q8. A beam of light of wavelength 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 the first dark fringes on either side of the central bright fringe is
Solution
β = 𝜆𝐷/𝑑 = (00 × 10-9 × 2)/ (1 × 10-3) = 12 × 10-4 m So, distance between the first dark fringes on either side of the central bright fringe 𝑋 = 2β = 2 ×12 × 10-4 m = 24 × 10-4 m = 2.4 mm
β = 𝜆𝐷/𝑑 = (00 × 10-9 × 2)/ (1 × 10-3) = 12 × 10-4 m So, distance between the first dark fringes on either side of the central bright fringe 𝑋 = 2β = 2 ×12 × 10-4 m = 24 × 10-4 m = 2.4 mm
Q9. Frequency of wave is 6 × 1015𝐻𝑧. The wave is
Solution
Wave is 𝑢𝑣 rays
Wave is 𝑢𝑣 rays
Q10. Oil floating on water looks coloured due to interference of light. What should be the order of magnitude of thickness of oil layer in order that this effect may be observed?
Solution
Oil floating on water looks coloured only when thickness of oil layer=wavelength of light=10000Å
Oil floating on water looks coloured only when thickness of oil layer=wavelength of light=10000Å
