## Wave Optics Quiz 3

JEE Advanced Physics Syllabus can be referred by the IIT aspirants to get a detailed list of all topics that are important in cracking the entrance examination. JEE Advanced syllabus for Physics has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with day-to-day experiences. In comparison to the other two subjects, the syllabus of JEE Advanced for physics is developed in such a way so as to test the deep understanding and application of concepts. .

Q1. In YDSE, find the thickness of a glass slab (Î¼=1.5) which should be placed before the upper slit S1 so that the central maximum now lies at a point where 5th bright fringe was lying earlier (before inserting the slab). Wavelength of light used is 5000 Ã…
•  5x10-6mm
•  3×10-6mm
•  10×10-6mm
•  5×10-5m
Solution

Q2.Light from a source emitting two wavelengths Î»1 and Î»2 is allowed to fall on Young’s double-slit apparatus after filtering one of the wavelengths. The position of interference maxima is noted. When the filter is removed both the wavelengths are incident and it is found that maximum intensity is produced where the fourth maxima occurred previously. If the other wavelength is filtered, at the same location the third maxima is found. What is the ratio of wavelengths?
•
2 / 3
•
3 / 2
•
3 / 4
•
4 / 3
Solution

Q3.  If the distance between the first maxima and fifth minima of a double-slit pattern is 7 mm and the slits are separated by 0.15 mm with the screen 50 cm from the slits, then wavelength of the light used is
•   600 nm
•  525 nm
•  467 nm
•  420 nm
Solution

Q4. In a Young’s double-slit experiment, the slits are illuminated by monochromatic light. The entire set-up is immersed in pure water. Which of the following act restore the original fringe width?
•  Bringing the slits close together
•  Moving the screen away from the slit plane
•  Replacing the incident light by that of longer wavelength
•  Introducing a thin transparent slab in front of one of the slits
Solution

Q5.If one of the two slits of a Young’s double-slit experiment is painted so that it transmits half the light intensity as the second slit, then
•  The fringe system will altogether disappear
•  The bright fringes will become brighter and the dark fringes will become darker
•  Both dark and bright fringes will become darker
•  Dark fringes will become brighter and bright fringes darker
Solution

Q6. Figure shows two coherent sources S1 and S2 emitting wavelength Î». The separation S1S2=1.5Î» and S1 is ahead in phase by Ï€/2 relative to S2. Then, the maxima occur in direction Î¸ given by sin-1 of
(i) 0   (ii) 1/2
(iii) -1/6   (iv) -5/6
Correct options are
•  (ii), (iii) and (iv)
•  (i), (ii) and (iii)
• (i), (iii) and (iv)
•  All the above
Solution

Q7.Light is incident at an angle Ï• with the normal to a plane containing two slits of separation d. Select the expression that correctly describes the positions of the interference maxima in terms of the incoming angle Ï• and outgoing angle Î¸
•   sinÏ•+sinÎ¸=(m+
1 / 2
)
Î» / d
•   dsinÎ¸=mÎ»
•   sinÏ•-sinÎ¸=(m+1)
Î» / d
•   sinÏ•+sinÎ¸=m
Î» / d
Solution

Q8. In Young’s double-slit experiment d/D=10-4=10-4 (d=distnace between slits, D=distance of screen from the slits). At a point P on the screen, resulting intensity is equal to the intensity due to the individual slit I0. Then, the distance of point P from the central maximum is (Î»=6000 Ã…)
•  2 mm
•  T1 mm
•  0.5 mm
•  4 mm
Solution

Q9. 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
•  1.2 mm
•  1.2 cm
•  2.4 cm
•  2.4 mm
Solution

Q10. Two identical coherent sources are placed on a diameter of a circle of radius R at separation x(≪R) symmetrical about the center of the circle. The sources emit identical wavelength Î» each. The number of points on the circle of maximum intensity is (x=5Î»)
•  20
•  22
•  24
• 26
Solution

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