1. In Newton’s ring experiment, the diameter of the 5th ring is 0.30 cm and diameter of 15th the ring is 0.62 cm. Find the diameter of the 25th ring.

Answer:

Diameter of Newton’s 5th ring = 0.30 cm

Diameter of Newton’s 15th ring = 0.62 cm

Diameter of Newton’s 25th ring = ?

From Newton’s rings experiment we infer that

D2n+m − D2n = 4λmR

For the 5th and 15th rings we have

D215 − D25 = 4λ * 10 * R _______ (1) (m = 10)

For 15th and 25th rings

D225 − D215 = 4λ * 10 * R _______ (2) (m = 10)

We equate the two derivatives

Equation (2) = Equation (1)

D225 − D215 = D215 − D25

D225 = 2D215 – D25

Substituting the values into the equation

D225 = 2 * 0.62 * 0.62 – 0.3 * 0.3 =0.6788 cm2

D25 = 0.8239 cm

Diameter of [tex]25^{th}[/tex] Newton  Ring = 0.97 cm

Newton Rings is an experiment based on principle of  thin film interference

In Newton Rings Experiment the Diameter of  [tex]n^{th}[/tex] dark ring is given by equation (1)

[tex]\rm D_n= 2\sqrt{n\lambda R} ......(1)\\where \; \\D_n = Diameter\; of \; n^{th} \; dark \; ring }\\n = Number \; of \; Ring\\\lambda = Wavelength \\R = Radius \; of \; Curvature \; of\; the \; lens[/tex]

From the condition given

[tex]\rm D_5 = 0.3 \; cm \\D_{15} = 0.62 \; cm\\\\D_{25} = To \; be \; determined \\[/tex]

Putting the values in equation (1) for fifth diameter we get

[tex]\rm D_5 = 0.3=2\sqrt{5\lambda R}.......(2) \\D_{15} = 0.62 = 2\sqrt{15\lambda R}.......(3) \\\\Equation \; (3) - Equation (2) \\\\0.32 = 2\sqrt{\lambda R} ( \sqrt{15} -\sqrt{5})\\\\2\sqrt{\lambda R } = 0.1954....(4)\\[/tex]

So  From equation (1) and (4)

[tex]\rm Diameter \; of \; 25^{th} Ring =D_{25} = 2\sqrt{\lambda R } \times \sqrt{25} \\\\D_{25} = 0.1954\times 5 \\\\D_{25} = 0.97 \; cm[/tex]

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