If space between lens and glass is filled with a liquid of refractive index μ = 1.25 and light of wavelength 5000 A ˚ is used, then what will be diameter (in mm) of 1 0 th dark ring. The radius of curvature of curved surface of the lens is 1m.

Dark ring radius: r n 2 = μ nλ R r 10 2 = 1.25 10 × 5000 × 1 0 − 10 × 1 = 4 × 1 0 − 6 , so r = 2 mm Diameter = 4 mm Answer 4.00

Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsThin film interferenceMedium2 minQB

PhysicsMediumnumerical

Passage / Comprehension

The Newton's ring experiment involves placing a convex lens on a flat glass plate and shining monochromatic light on lens. A concave shaped air film is formed between the lens and plate. The light incident from top reflects from bottom surface of the top lens and the top surface of the flat plate.\\

\\ The result is a series of concentric rings that alternate between bright and dark, with a dark spot at the center.

If space between lens and glass is filled with a liquid of refractive index $μ = 1.25$ and light of wavelength $5000 \overset{˚}{A}$ is used, then what will be diameter (in mm) of $1 0^{th}$ dark ring. The radius of curvature of curved surface of the lens is 1m.

Answer:

4.00

Solution:

Dark ring radius: $r_{n}^{2} = \frac{nλ R}{μ}$

$r_{10}^{2} = \frac{10 \times 5000 \times 1 0 ^{- 10} \times 1}{1.25} = 4 \times 1 0^{- 6}$ , so $r = 2 mm$

Diameter = 4 mm

Answer 4.00

Stream:JEESubject:PhysicsTopic:Wave OpticsSubtopic:Thin film interference

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...