Exams › JEE Main › Physics

A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on a screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4th bright fringe of the unknown light. From this data, the wavelength of the unknown light is:

1. 885.0 nm
2. 442.5 nm
3. 776.8 nm
4. 393.4 nm

Correct answer: 442.5 nm

Solution

The condition for the bright fringes in Young's double slit experiment is given by the formula mλ = d sin(θ), where m is the order of the fringe, λ is the wavelength, and d is the distance between the slits. Since the third bright fringe of the known light (590 nm) coincides with the fourth bright fringe of the unknown light, we can set up the equation 3(590 nm) = 4(λ_unknown). Solving this gives λ_unknown = 442.5 nm.

Related JEE Main Physics questions

• The pupil opening of a human eye is about 2 mm. If the average wavelength of visible light is taken as 5000 Å, the eye’s approximate angular resolving limit is
• For a single-slit diffraction pattern, the condition corresponding to the appearance of secondary maxima is
• In a Fresnel biprism setup, the two lens positions give slit separations of 16 cm and 9 cm. What is the true separation between the slits?
• In a Young’s double-slit setup, two monochromatic sources emit light of wavelengths 2500 Å and 3500 Å. Which pair of fringe orders from the two patterns will fall at the same position?
• In a Young’s double-slit setup, the intensity at a bright fringe maximum is I0. The slit separation is d = 5bb, where bb is the wavelength of the light used. If the screen is placed at a distance D = 10d, what intensity is observed at the point directly opposite one of the slits?
• What is the relationship between the polarizing angle and the critical angle?

⚔️ Practice JEE Main Physics free + battle 1v1 →