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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.  As a result of interference of two coherent sources of light energy is
•   Redistributed and the distribution does not very with time
•   Increased
•   Redistributed and that distribution changes with time
•   Decreased
Solution
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
•   A bright slit will be observed, no interference pattern will exist
•   The bright fringes will become more bright
•   The bright fringes will become fainter
•   None of the above
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

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
•   5 𝑐𝑚
•   2.5 𝑐𝑚
•   0.25 𝑐𝑚
•   0.1 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

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
•   1.2cm
•   1.2mm
•   2.4cm
•   2.4mm
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

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
•   6000 Å
•   5000 Å
•   3000 Å
•   1500 Å
Solution
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
•   400 𝑛𝑚 and 450 𝑛𝑚
•   425 𝑛𝑚 and 400 𝑛𝑚
• 400 𝑛𝑚 and 425 𝑛𝑚
•   450 𝑛𝑚 and 400 𝑛𝑚
Solution

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

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
•   Increase of 589 Å
• Decrease of 589 Å
•   Increase of 6479 Å
•   Zero
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

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

•   That the central maxima is narrower
•   No diffraction pattern
•   More number of fringes
•   Less number of fringes
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

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
•   0.24 cm
•   3.2 cm
•   1.28 cm
• 0.32 cm
Solution
Distance between successive fringes-fringe width = 𝛽 = 𝜆𝐷/𝑑 = 8×10-5×2 /0.05 = 0.32 cm ## Want to know more

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE: WAVE OPTICS QUIZ 18
WAVE OPTICS QUIZ 18