Question

Deduce an expression for diameter of the n dark ring in newtons ring

​

Answer 1

Central dark spot: At the point of contact of the lens with the glass plate the thickness of the air film is very small compared to the wavelength of light therefore the path difference introduced between the interfering waves is zero.

radhe radhe ❤️

Answer 2

Answer:

The radius of the dark king is directly proportional to square root of its order. i.e $r_{n} =$√nRλ

Explanation:

Expression for the radius of the nth dark ring can be obtained as follows :

• Consider the Plano convex lens vertical section SOP taken via its curvature center C.
• Let O represent the lens's point of contact with the plane surface and R represent the planoconvex lens radius of curvature.
• Let t represent the air film's thickness at S and P. Draw ST and PQ parallel to the glass plate's planar surface.

Therefore, ST = AO = PQ = t.

Let $r_{n}$ be the radius of the $n^{th}$ dark ring that passes through the points S and P.

SA = AP = $r_{n}$

The vertical diameter of the circle is ON  then by law of segments is given as

SA. AP = OA .AN

$r^{2} = t(2R - t)$

$r^{2} = 2Rt- t^{2}$

Now neglect the $t^{2}$ term comparing with 2R

we get,

$2t = \frac{r^{2} _{n} }{R}$

According to the condition for darkness

2t = nλ

$\frac{r^{2}_{n} }{R}$ = nλ

⇒ $r^{2} _{n} =$ nRλ

$r_{n} =$√nRλ

R and λ are constants,

Therefore radius of the dark king is directly proportional to square root of its order.

#SPJ2