WAVE OPTICS QUIZ 16

 

 WAVE OPTICS Quiz-16

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. .

Q1.  . A radio receiver antenna that is 2 𝑚 long is oriented along the direction of the electromagnetic wave and receives a signal of intensity 5 × 10^-16𝑊/𝑚². The maximum instantaneous potential difference across the two ends of the antenna is

•   1.23 𝜇𝑉
•   1.23 𝑚𝑉
•   1.23 𝑉
•   12.3 𝑚𝑉

Solution
𝐼 = 1/2ε₀𝐶𝐸₀² 

 ⇒ 𝐸_o = √2𝐼ε₀𝑐 = √ (2 × 5 × 10^-16 )/ (8.85 × 10^-12 × 3 × 10⁸)

 = 0.61 × 10^-6 𝑉 /𝑚 Also 𝐸₀ = 𝑉₀/ 𝑑 ⇒ 𝑉₀= 𝐸₀𝑑 = 0.61 × 10_-6 × 2 = 1.23𝜇𝑉

Q2. The two slits are 1𝑚𝑚 apart from each other and illuminated with a light of wavelength 5 × 10^-7𝑚. If the distance of the screen is 1 𝑚 from the slits, then the distance between third dark fringe and fifth bright fringe is

•   1.5 𝑚𝑚
•   0.75 𝑚𝑚
•   1.25 𝑚𝑚
•   0.625 𝑚𝑚

Solution
Distance of 5th bright fringe from central fringe, 𝑋_5D = 5𝜆𝐷/ 𝑑 …(i) 

Distance of 3rd dark fringe from central fringe, 𝑋_3D = (2 × 3 − 1)𝜆𝐷 /2𝑑 = 5 /2 𝜆𝐷 /𝑑 …(ii) 

 From (i) and (ii) required distance 

 𝑋_5B − 𝑋_3D = (5 − 5 /2 ) 𝜆𝐷 /𝑑 = 5 /2 × (5 × 10^-7 × 1 )/(1 × 10^-3) = 1.25 𝑚𝑚

Q3.   . In Young’s double slit experiment, distance between two sources is 0.1 mm. The distance of screen from the source is 20 cm. Wavelength of light used is 5460 Åhen angular position of first dark fringe is

•   0.08°
•   0.16°
•   0.20°
•   0.32°

Solution
𝑑 = 0.1 mm = 10^-4,D = 20 cm = 1/5 m 

 𝜆 = 5460Å = 5.46 × 10^-7m 

 Angular position of first dark fringe is

 θ = 𝑥/𝐷 = 𝜆 /2𝑑 = 5.46 × 10^-7 / 2 × 10^-4 

= 2.73 × 10^-3 rad = 2.73 × 10^-3 × 180°/𝜋 = 0.156°

Q4.  . The maximum distance upto which TV transmission from a TV tower of height ℎ can be received is proportional to

•   ℎ^1/2
•   h
•   h
•   h²

Solution
Distance covered by T.V. signals = √2ℎ𝑅 ⇒ maximum distance ∝ ℎ1/2

Q5. In Young’s double slit experiment, the intensity on the screen at a point where path difference 𝜆 is 𝐾. What will be the intensity at the point where path difference is 𝜆/4

•   K/4
•   K/2
•   K
•   ZERO

Solution
  By using phase difference 𝜙 = 2𝜋/𝜆 (Δ) 

For path difference 𝜆, phase difference 𝜙₁ = 2𝜋 and for path difference 𝜆/4, phase difference 𝜙₂ = 𝜋/2 

Also by using 𝐼 = 4𝐼0 cos2 𝜙/2 ⇒ 𝐼1 /𝐼2 = cos²(𝜙1/2) cos²(𝜙2/2) 

 ⇒ 𝐾/ 𝐼₂ = cos²(2𝜋/2)/ cos² (𝜋/2 2 ) = 1 /1/2 ⇒ 𝐼₂ = 𝐾/ 2 .

Q6.  In Young’s double slit experiment, the separation between the slit is haled and the distance between the slits and screen is doubled. The fringe-width will

•   Be halved
•   Be doubled
• Be quadrupled
•   Remain unchanged

Solution
Let S be a slit illuminated by monochromatic light of wavelength(𝜆), let 𝑆₁,𝑆₂ be coherent sources and distance between them be d and distance between source and screen is D. then, fringe width ( W ) is given by 

 𝑊 = 𝐷𝜆/𝑑 

When, 𝑑₂ = 𝑑/2 ,𝐷₂ = 2𝐷 

 ∴ 𝑊₂ = (2𝐷)𝜆/(𝑑/2) = 4 𝐷𝜆/ 𝑑 = 4𝑊 

 The fringe width is quadrupled.

Q7. The distance between the first dark and bright band formed in Young’s double slit experiment with band width B is

•   B/4
•   B
•   B/2
•   3B/2

Solution

Position of nth bright fringe 𝑥₁ = 𝑛𝜆𝐷 /𝑑 

For first bright fringe 𝑛 = 1 ∴ 𝑥₁ = 𝜆𝐷/ 𝑑

 Position of nth dark fringe 𝑥₂ = (2𝑛−1)𝜆𝐷/ 2𝑑

 For first dark fringe 𝑛 = 1 ∴ 𝑥₂ = 𝜆𝐷 /2𝑑 

Now, 𝑥₁ − 𝑥₂ = 𝜆𝐷/ 2𝑑 If B is the band width, then 𝑥₁ − 𝑥₂ =𝐵 /2

Q8. Which of the following statements indicates that light waves are transverse

•   Light waves can travel in vacuum
• Light waves show interference
•   Light waves can be polarized
•   Light waves can be diffracted

Solution
Transverse waves can be polarized only

Q9. In Young’s double slit experiment, the central bright fringe can be identified

•   As it has greater intensity than the other bright fringes
•   As it is wider than the other bright fringes
•   As it is narrower than the other bright fringes
•   By using white light instead of monochromatic light

Solution
When white light is used instead of monochromatic light, the central bright fringe becomes white, while others are coloured. Hence, distinction is made

Q10.  The observed wavelength of light coming from a distant galaxy is found to be increased by 0.5% as compared with that coming from a terrestrial source. The galaxy is

•   Stationary with respect to the earth
•   Approaching the earth with velocity of light
•   Receding from the earth with the velocity of light
• Receding from the earth with a velocity equal to 1.5 × 10⁶𝑚/𝑠

Solution
Δ𝜆/𝜆 = 𝑣 /𝑐 , Now Δ𝜆 = 0.5/ 100 𝜆 ⇒ Δ𝜆 /𝜆 = 0.5 /100

 ∴ 𝑣 = 0.5/ 100 × 𝑐 = 0.5/ 100 × 3 × 108 = 1.5 × 10⁶𝑚/𝑠 

 Increase in 𝜆 indicates that the star is receding