Exams › JEE Advanced › Physics

In a Young's double-slit experiment the slit separation is 0.5 mm and the interference pattern is seen on a screen 100 cm from the slits. The 9th bright fringe is found 7.5 mm from the centre of the pattern (on one side). What is the wavelength of the light used?

1. 4167 angstrom
2. 5000 angstrom
3. 6000 angstrom
4. 3750 angstrom

Correct answer: 4167 angstrom

Solution

Bright fringes occur at y = n*lambda*D/d. Substituting the 9th fringe position gives the wavelength directly.

Related JEE Advanced Physics questions

• What factor influences the visibility of fringes in an interference pattern?
• When the screen is moved further along the x-axis away from the source, which of the following statements is false?
• In the Young's double slit experiment using a monochromatic light of wavelength λ, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is
• Using the expression 2d sin θ = λ, one calculates the values of d by measuring the corresponding angles θ in the range 0 to 90°. The wavelength λ is exactly known and the error in θ is constant for all values of θ. As θ increases from 0°:
• Two coherent point sources separated by a distance d = 5 micrometers emit light of wavelength lambda = 2 micrometers in phase. A circular wire of radius R = 20 micrometers is placed symmetrically around the two sources (sources lie along a diameter). Points A and B are at the ends of the diameter along the line joining the sources; points C and D are at the ends of the diameter perpendicular to the line joining the sources. Which of the following is correct?
• Young's double slit experiment is performed in a medium of refractive index mu1 = 4/3. A thin transparent slab of refractive index mu2 = 3/2 and thickness t = 8 micrometers is placed in front of slit S2. The magnitude of the optical path difference (with respect to the liquid medium) at the central point O is:

⚔️ Practice JEE Advanced Physics free + battle 1v1 →