Question

In ydse how many maxima can be obtained on the screen if wavelength of light used is 200nm and d= 700nm

Answer 1

Let y be wavelength, Ф be angular width of a fringe and d be distance between two slits.

Ф = y / d

Therefore maximum angle can be 90° hence number of fringes:

(n) = Sin 90° / Sin Ф

As Sin Ф is very small Sin Ф ≈ Ф

n = 1 / Ф

n = d / y

700 / 200 = 7/2

Number of maxima or minima = 2n + 1

2 × 7/2 + 1 = 8

The answer is thus : 8

Answer 2

Answer:

The number of maxima is 8.

Explanation:

Given that,

Wave length =200 nm

Distance d = 700 nm

The fringe width is

$\theta=\dfrac{\lambda}{d}$.....(I)

Where, $\theta$ =angular width of fringe

d = distance between the two slits

$\lambda$=wave length of the light

Put the value of wave length and distance in equation (I)

$\theta = \dfrac{200}{700}$

$\theta=\dfrac{2}{7}$

The maximum angle can be 90°

Now, The number of fringe is

$n = \dfrac{\sin\ 90^{\circ}}{\sin\theta}$

Here, sin Ф is very small so sin Ф≈Ф

Therefore,

$n = \dfrac{1}{\theta}$...(II)

Put the value of Ф in equation (II)

$n = \dfrac{7}{2}$

The number of maxima and minima is

$2n-1$

Put the value of n in equation (III)

$2\times\dfrac{7}{2}+1=8$

Hence, The number of maxima is 8.