Exams › JEE Main › Physics

Using light of wavelength 6000 angstrom, the smallest separation between two points that a microscope can just resolve is 0.1 mm. If the wavelength is changed to 4800 angstrom, what does this resolution limit become?

1. 0.80 mm
2. 0.12 mm
3. 0.10 mm
4. 0.08 mm

Correct answer: 0.08 mm

Solution

Resolving limit d is proportional to lambda (d = 1.22*lambda/(2*n*sin theta)). So d2 = d1*(lambda2/lambda1) = 0.1 mm * (4800/6000) = 0.1 * 0.8 = 0.08 mm.

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 →