Exams › JEE Main › Physics

Monochromatic light of wavelength 6000 angstrom falls normally on a single slit of width 24 x 10⁻⁵ cm. Determine the angular position of the second diffraction minimum measured from the central maximum.

1. 15 deg
2. 30 deg
3. 60 deg
4. 74 deg

Correct answer: 30 deg

Solution

With a = 24x10⁻⁵ cm = 2.4x10⁻⁶ m and lambda = 6000 angstrom = 6x10⁻⁷ m, the second minimum (n=2) satisfies sin(theta) = 2 lambda / a = 2(6x10⁻⁷)/(2.4x10⁻⁶) = 0.5, so theta = 30 deg.

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 →