Question

In Newton's ring experiment, the diameter of 15th ring was found to be 0.6cm, and that of the 5th ring is 0.346cm. If the radius of curvature is 100 cm, then the wavelength of light

Answer 1

Explanation:

Solution : using formula, \lambda=\frac{D_{n+m}^2-D_n^2}{4mR}λ=

4mR

D

n+m

2

−D

n

2

Here diameter of 15th ring, D_{n+m}D

n+m

= 0.59 cm = 5.9 × 10¯³ m

diameter of 5th ring, D_nD

n

= 0.336 cm = 3.36 × 10¯³ m

Here, m = 15 - 5 = 10

R = 100 cm = 1 m

So, wavelength, λ = [(5.9 × 10¯³)² - (3.36 × 10¯³)²]/4 × 10 × 1

= {(34.81 - 11.2896)/40} × 10¯⁶

= 23.5204/40 × 10¯⁶

= 0.5880 × 10¯⁶ m

= 5880 A°