Exams › JEE Main › Physics

In a Fresnel biprism experiment the source-to-screen distance is 1 m and the source-to-biprism distance is 10 cm. Light of wavelength 6000 angstrom is used, the measured fringe width is 0.03 cm, and the refracting angle of the biprism is 1 deg. Find the refractive index of the biprism material.

1. 1.57
2. 1.33
3. 1.50
4. 1.68

Correct answer: 1.57

Solution

From beta = lambda*D/d, find d = 2*10⁻³ m. Then using d = 2*a*(mu-1)*alpha with a = 0.10 m and alpha = 1 deg (0.01745 rad) gives (mu - 1) ~ 0.573, so mu ~ 1.57.

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 →