WAVE OPTICS Quiz-13
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. A point source of electromagnetic radiation has an average power output of 800 𝑊. The maximum value of electric field at a distance 4.0 𝑚 from the source is
Solution
Intensity of EM wave is given by 𝐼 = 𝑃/ 4𝜋𝑅2 = 𝑣𝑎𝑣.𝑐 = 1/ 2 εo𝐸02 × 𝑐 ⇒ 𝐸0 = √ 𝑃/ 2𝜋𝑅2𝑐
= √
800/ (2 × 3.14 × (4)2 × 8.85 × 10-12 × 3 × 108)
= 54.77
𝑉/𝑚
Intensity of EM wave is given by 𝐼 = 𝑃/ 4𝜋𝑅2 = 𝑣𝑎𝑣.𝑐 = 1/ 2 εo𝐸02 × 𝑐 ⇒ 𝐸0 = √ 𝑃/ 2𝜋𝑅2
Q2.. In an electromagnetic wave, the amplitude of electric field is 1 𝑉/𝑚, the frequency of wave is 5 × 1014𝐻𝑧. The wave is propagating along 𝑧-axis. The average energy density of electric field, in 𝐽𝑜𝑢𝑙𝑒/𝑚3, will be
Solution
Average energy density of electric field is given by 𝑢e = 1/ 2 εo 𝐸2 = 1/ 2 εo( 𝐸0 /√2 )2 = 1/ 4 εo𝐸02 = 1 /4 ×8.85 × 10-12(1)2 = 2.2 × 10-12𝐽/𝑚3
Average energy density of electric field is given by 𝑢e = 1/ 2 εo 𝐸2 = 1/ 2 εo( 𝐸0 /√2 )2 = 1/ 4 εo𝐸02 = 1 /4 ×8.85 × 10-12(1)2 = 2.2 × 10-12𝐽/𝑚3
Q3. In the Young’s double slit experiment, the spacing between two slits is 0.1 𝑚𝑚. If the screen is kept at a distance of 1.0 𝑚 from the slits and wavelength of light is 5000 Å, then the fringe width is
Solution
𝛽 = 𝜆𝐷 𝑑 = (500 × 10-10 × 1) / (0.1 × 10-3) 𝑚 = 5 × 10-3𝑚 = 0.5 𝑐𝑚
𝛽 = 𝜆𝐷 𝑑 = (500 × 10-10 × 1) / (0.1 × 10-3) 𝑚 = 5 × 10-3𝑚 = 0.5 𝑐𝑚
Q4. Which of the following diagrams represent the variation of electric field vector with time for a circularly polarized light
Solution
To form circularly polarized light 𝐸x = 𝐴sin𝜔𝑡 𝐸y = 𝐴cos𝜔𝑡 Resultant amplitude |𝐸 ⃗ |2 = 𝐴2 + 𝐴2 + 2𝐴.𝐴cos𝜋/ 2 ⇒ |𝐸 ⃗ | = 𝐴√2 = constant .
To form circularly polarized light 𝐸x = 𝐴sin𝜔𝑡 𝐸y = 𝐴cos𝜔𝑡 Resultant amplitude |𝐸 ⃗ |2 = 𝐴2 + 𝐴2 + 2𝐴.𝐴cos𝜋/ 2 ⇒ |𝐸 ⃗ | = 𝐴√2 = constant .
Q5. In a diffraction pattern by a wire, on increasing diameter of wire, fringe width
Solution
𝛽 = 𝜆.𝐷/ 𝑑 where 𝐷 = distance of screen from wire, 𝑑 = diameter of wire .
𝛽 = 𝜆.𝐷/ 𝑑 where 𝐷 = distance of screen from wire, 𝑑 = diameter of wire .
Q6. Light if wavelength 2 × 10-3 m falls on a slit of width 4 × 10-3 m. The angular dispersion of the central maximum will be
Solution
Angular dispersion of central maximum=angular dispersion of 1st minimum (= 2θ) From sinθ = 1𝜆/𝑎 = (2×10-3)/ (4×10-3) = 1/2 θ = 30° ∴ 2θ = 2 × 30° = 60°
Angular dispersion of central maximum=angular dispersion of 1st minimum (= 2θ) From sinθ = 1𝜆/𝑎 = (2×10-3)/ (4×10-3) = 1/2 θ = 30° ∴ 2θ = 2 × 30° = 60°
Q7. . The intensity of gamma radiation from a given source is 𝐼. On passing through 36 𝑚𝑚 of lead, it is reduced to 𝐼/8 . The thickness of lead which will reduce the intensity to 𝐼/2 will be
Solution
𝐼′ = 𝐼 𝑒−𝜇𝑥 ⇒ 𝑥 = 1 𝜇 log𝑒 𝐼/𝐼′ (where 𝐼 = original intensity, 𝐼′ = changed intensity) 36 = 1/𝜇 loge 𝐼 /(𝐼/8) = 3/𝜇 loge 2 …(i) 𝑥 = 1 𝜇 loge 𝐼 𝐼/2 = 1 𝜇 loge 2 …(ii) From equation (i) and (ii), 𝑥 = 12𝑚𝑚
𝐼′ = 𝐼 𝑒−𝜇𝑥 ⇒ 𝑥 = 1 𝜇 log𝑒 𝐼/𝐼′ (where 𝐼 = original intensity, 𝐼′ = changed intensity) 36 = 1/𝜇 loge 𝐼 /(𝐼/8) = 3/𝜇 loge 2 …(i) 𝑥 = 1 𝜇 loge 𝐼 𝐼/2 = 1 𝜇 loge 2 …(ii) From equation (i) and (ii), 𝑥 = 12𝑚𝑚
Q8. . In the Young’s double slit experiment, a mica slip of thickness 𝑡 and refractive index 𝜇 is introduced in the ray from first source𝑆1. By how much distance fringes pattern will be displaced.
Solution
For a path difference (𝜇 − 1)𝑡, the shift is x = (μ − 1)t D/d
For a path difference (𝜇 − 1)𝑡, the shift is x = (μ − 1)t D/d
Q9.. In a biprism experiment, by using light of wavelength5000 Å, 5mm wide fringes are obtained on a screen 1.0 m away from the coherent sources. The separation between the two coherent sources is
Solution
From 𝛽 = 𝜆𝐷/𝑑 5 ×10-3 = ((5000 × 10−10) × 1.0 )/𝑑 𝑑 = 5 × 10-7 / 5 × 10-3 = 10-4m = 0.1 mm
From 𝛽 = 𝜆𝐷/𝑑 5 ×10-3 = ((5000 × 10−10) × 1.0 )/𝑑 𝑑 = 5 × 10-7 / 5 × 10-3 = 10-4m = 0.1 mm
Q10. In Young’s double slit experiment, an interference pattern is obtained on a screen by a light of wavelength 6000Å coming from the coherent sources 𝑆1 and 𝑆2. At certain point P on the screen third dark fringe is formed. Then the path difference 𝑆1𝑃 − 𝑆2𝑃 in microns is
Solution
𝜆 = 6000 Å = 6 × 10-7 m Path difference for dark fringe ∆𝑥 = (2𝑛 + 1) 𝜆/2 For third dark fringe 𝑛 = 2 ∴ ∆𝑥 = (2 × 2 + 1) × (6 × 10-7) /2 = 5 ×6 × 10-7 /2 = 15 × 10-7 = 1.5 × 10-6 m = 1.5𝜇
𝜆 = 6000 Å = 6 × 10-7 m Path difference for dark fringe ∆𝑥 = (2𝑛 + 1) 𝜆/2 For third dark fringe 𝑛 = 2 ∴ ∆𝑥 = (2 × 2 + 1) × (6 × 10-7) /2 = 5 ×6 × 10-7 /2 = 15 × 10-7 = 1.5 × 10-6 m = 1.5𝜇
