WAVE OPTICS Quiz-18
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. As a result of interference of two coherent sources of light energy is
Solution
Redistributed and the distribution does not very with time
Redistributed and the distribution does not very with time
Q2. In Young’s double slit experiment, if one of the slits is closed fully, then in the interference pattern
Solution
If one of slits is closed then interference fringes are not formed on the screen but a fringe pattern is observed due to diffraction from slit
If one of slits is closed then interference fringes are not formed on the screen but a fringe pattern is observed due to diffraction from slit
Q3. If a transparent medium of refractive index 𝜇 = 1.5 and thickness 𝑡 = 2.5 × 10-5𝑚 is inserted in front of one of the slits of Young’s Double Slit experiment, how much will be the shift in the interference pattern? The distance between the slits is 0.5 𝑚𝑚 and that between slits and screen is 100 𝑐𝑚
Solution
Shift in the fringe pattern 𝑥 = (𝜇−1)𝑡.𝐷 /𝑑 = (1.5 − 1) × 2.5 × 10-5 × 100 × 10-2 / 0.5 × 10-3 = 2.5 𝑐𝑚
Shift in the fringe pattern 𝑥 = (𝜇−1)𝑡.𝐷 /𝑑 = (1.5 − 1) × 2.5 × 10-5 × 100 × 10-2 / 0.5 × 10-3 = 2.5 𝑐𝑚
Q4. A beam of light of wavelength 600 nm from a distance source falls on a single slit 1.00 mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of the central bright fringe is
Solution
𝜆 = 600 nm = 6 × 10-7m 𝑎 = 1 mm = 10-3m,𝐷 = 2 m Distance between the first dark fringes on either side of central bright fringe=width of central maximum = 2𝜆𝐷/𝑎 = (2 × 6 × 10-7 × 2) / 10-3 = 24 × 10-4m = 2.4 mm
𝜆 = 600 nm = 6 × 10-7m 𝑎 = 1 mm = 10-3m,𝐷 = 2 m Distance between the first dark fringes on either side of central bright fringe=width of central maximum = 2𝜆𝐷/𝑎 = (2 × 6 × 10-7 × 2) / 10-3 = 24 × 10-4m = 2.4 mm
Q5. A light wave is incident normally over a slit of width 24 × 10-5𝑐𝑚. The angular position of second dark fringe from the central maxima is 30°. What is the wavelength of light
Solution
For second dark fringe 𝑑sin𝜃 = 2𝜆 ⇒ 24 × 10-5 × 10-2× sin30 = 2𝜆 ⇒ 𝜆 = 6 × 10-7𝑚 = 6000 Å
For second dark fringe 𝑑sin𝜃 = 2𝜆 ⇒ 24 × 10-5 × 10-2× sin30 = 2𝜆 ⇒ 𝜆 = 6 × 10-7𝑚 = 6000 Å
Q6. If the eighth bright band due to light of wavelength 𝜆1 coincides with ninth bright band from light of wavelength 𝜆2 in Young’s double slit experiment, then the possible wavelength of visible light are
Solution
450 𝑛𝑚 and 400 𝑛𝑚
450 𝑛𝑚 and 400 𝑛𝑚
Q7. Light of wavelength 6000 Å falls on a single slit of width 0.1 mm. The second minimum will be formed for the angle of diffraction of
Solution
Given single slit of width 𝑑 = 0.1 mm 𝑑 = 0.1 × 10-3 m Or 𝑑 = 1 × 10-4m Light of wavelength 𝑎 = 600 Å Or 𝛼 = 6 × 10-7 m The angle of diffraction θ = 𝑛𝜆/𝑑 θ = 2 × 6 × 10-7 / 1 × 10-4 θ = 12 × 10-3 θ = 0.012 ra
Given single slit of width 𝑑 = 0.1 mm 𝑑 = 0.1 × 10-3 m Or 𝑑 = 1 × 10-4m Light of wavelength 𝑎 = 600 Å Or 𝛼 = 6 × 10-7 m The angle of diffraction θ = 𝑛𝜆/𝑑 θ = 2 × 6 × 10-7 / 1 × 10-4 θ = 12 × 10-3 θ = 0.012 ra
Q8. In double slit experiment, the angular width of the fringes is 0.20° for the sodium light (𝜆 = 5890Å). In Order to increase the angular width of the fringes by 10%, the necessary change in the wavelength is
Solution
𝜃 = 𝜆/𝑑;𝜃 can be increased by increasing 𝜆, so here 𝜆 has to be increased by 10% 𝑖.𝑒.,% Increase = 10 /100 × 5890 = 589Å
𝜃 = 𝜆/𝑑;𝜃 can be increased by increasing 𝜆, so here 𝜆 has to be increased by 10% 𝑖.𝑒.,% Increase = 10 /100 × 5890 = 589Å
Q9. Yellow light is used in single slit diffraction experiment with slit width 0.6 𝑚𝑚. If yellow light is replaced by 𝑋- rays then the pattern will reveal
Solution
Diffraction is obtained when the slit width is of the order of wavelength of EM waves (or light). Here wavelength of 𝑋-rays (1 − 100Å) is veryvery lesser than slit width (0.6 𝑚𝑚). Therefore no diffraction pattern will be observed
Diffraction is obtained when the slit width is of the order of wavelength of EM waves (or light). Here wavelength of 𝑋-rays (1 − 100Å) is veryvery lesser than slit width (0.6 𝑚𝑚). Therefore no diffraction pattern will be observed
Q10. In a double slit interference experiment, the distance between the slits is 0.05 cm and screen is 2 m away from the slits. The wavelength of light is 8.0 × 10-5 cm. The distance between successive fringes is
Solution
Distance between successive fringes-fringe width = 𝛽 = 𝜆𝐷/𝑑 = 8×10-5×2 /0.05 = 0.32 cm
Distance between successive fringes-fringe width = 𝛽 = 𝜆𝐷/𝑑 = 8×10-5×2 /0.05 = 0.32 cm
