Exams › JEE Advanced › Physics

An experimenter uses a Michelson interferometer to measure the wavelength of light. While slowly moving mirror M2, 10000 new bright fringes appear at the centre. The mirror moves a total distance of 3.164 mm. What is the wavelength of the light in nm?

1. 316.4
2. 632.8
3. 158.2
4. 1265.6

Correct answer: 632.8

Solution

lambda = 2*d/N = 2*3.164*10⁻³ / 10000 = 6.328*10⁻⁷ m = 632.8 nm.

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 →