Fringe Width and Conditions for Maxima Minima - UNSOLVED PRACTICE SET
Chapter: Wave Optics | Topic: Fringe Width and Conditions for Maxima Minima
FRINGE WIDTH AND CONDITIONS FOR MAXIMA MINIMA - UNSOLVED PRACTICE SET
Topic: Fringe Width and Conditions for Maxima Minima
Multiple Choice Questions
Q1. The fringe width in YDSE is directly proportional to:
- The slit separation
- The wavelength of light
- The inverse of the distance to the screen
- The square of the slit separation
Q2. For constructive interference (bright fringe), the path difference must be:
- An odd multiple of Ī»/2
- An integral multiple of Ī»
- Zero only
- An integral multiple of Ī»/2
Q3. For destructive interference (dark fringe), the path difference must be:
- An integral multiple of Ī»
- An odd multiple of Ī»/2
- Zero
- An even multiple of Ī»
Q4. If the wavelength of light used in YDSE is doubled, the fringe width:
- Remains the same
- Becomes half
- Doubles
- Becomes four times
Q5. The angular fringe width is given by:
- Īø = Ī»/d
- Īø = Ī»D/d
- Īø = d/Ī»
- Īø = D/d
Q6. The position of the nth bright fringe from the central fringe is:
- y_n = nĪ»D/(2d)
- y_n = nĪ»D/d
- y_n = (2n + 1)Ī»D/(2d)
- y_n = (2n ā 1)Ī»D/(2d)
Short Answer Questions
Q7. Define fringe width. On what factors does it depend?
Q8. Write the conditions for constructive and destructive interference in YDSE in terms of path difference.
Q9. Why does the fringe width decrease when the slit separation is increased?
Q10. In YDSE, the 3rd bright fringe is at a distance of 3 mm from the central fringe. Where will the 6th bright fringe be located?
Q11. What is angular fringe width? How is it related to the linear fringe width?
Q12. If the entire YDSE apparatus is immersed in a liquid of refractive index n, how does the fringe width change? Explain.
Long Answer Questions
Q13. Derive the expression for fringe width in Young's double slit experiment. Show that it is directly proportional to the wavelength and the distance to the screen, and inversely proportional to the slit separation.
Q14. Derive the conditions for maxima and minima in YDSE. Show that bright fringes occur when the path difference is nĪ» and dark fringes occur when the path difference is (2n + 1)Ī»/2.
Q15. Explain how the fringe pattern changes when:
(a) The experiment is performed in a medium other than air
(b) White light is used instead of monochromatic light
(c) One slit is covered with a thin transparent sheet
Numerical & Application-Based Problems
Q16. In YDSE, the slit separation is 0.3 mm and the distance to the screen is 1.8 m. Light of wavelength 500 nm is used.
(a) Calculate the fringe width.
(b) Calculate the distance between the 4th bright fringe and the 2nd dark fringe on the same side of the central fringe.
(c) Calculate the angular position of the 3rd dark fringe.
(d) If the screen is moved 0.5 m closer to the slits, calculate the new fringe width.
Q17. In a YDSE setup, the fringe width is measured as 0.8 mm. When the wavelength is changed to 600 nm, the fringe width becomes 1.0 mm.
(a) Calculate the ratio of the two wavelengths.
(b) If the distance to the screen is 2 m, calculate the slit separation.
(c) Calculate the position of the 5th dark fringe for the second wavelength.
Q18. In your school's physics lab, a student performs YDSE and records the following data:
Slit separation: 0.5 mm
Distance to screen: 2.0 m
Wavelength: 550 nm
(a) Calculate the expected fringe width.
(b) The student measures the distance between the 2nd bright fringe on one side and the 2nd bright fringe on the other side as 8.8 mm. Calculate the experimental fringe width and the percentage error.
(c) The student then places a thin mica sheet (n = 1.6) in front of one slit and observes that the central fringe shifts to the position previously occupied by the 4th bright fringe. Calculate the thickness of the mica sheet.
(d) A classmate suggests that instead of measuring fringe width, they should measure the angular separation between fringes using a protractor. Discuss the feasibility and accuracy of this method.
(e) In an optics factory, YDSE is used to measure the wavelength of light from a new laser source. If the measured fringe width is 1.2 mm for d = 0.2 mm and D = 2.5 m, calculate the wavelength. Why is this method preferred over direct measurement of wavelength?