Question

the slit increased,the separation distance between two consecutive bright bonds will decreases.Explain?

Answer 1

Answer on Question #58352, Physics / Optics

the slit increased, the separation distance between two consecutive bright bonds will decreases. Explain?

Consider the task.

At the slit of width $a$ mm normally falls the parallel beam of monochromatic light of known wavelength $\lambda$ . Lens is placed behind slit. The diffraction pattern using a lens projected on the screen. The distance from the lens to the screen $L$ m. Determine the width of the peak in the diffraction pattern.

Find: $l - ?$

Given:

a

A

L

Solution:

This is the phenomenon of diffraction.

Central maximum takes a distance between the first two lows.

Conditions for diffraction minimum on slit:

$$a \sin \varphi = k \lambda (1),$$

where $k = \pm 1, \pm 2, \pm 3, \ldots$

Of $\Delta OAB$ (angle OAB = 90°) $\Rightarrow$ $\tan \varphi = \frac{1}{2L}$ (2)

Of (2) $\Rightarrow l = 2L \tan \varphi$ (3)

We consider the first order maxima. Therefore the diffraction angles $\varphi$ are small.

For small angles: $\sin \varphi \approx \tan \varphi$ (4)

Of (1) $\Rightarrow$ $\sin \varphi = \frac{k\lambda}{a}$ (5)

(5) in (4): $\tan \varphi = \frac{k\lambda}{a}$ (6)

(6) in (3): $l = \frac{2Lk\lambda}{a}$ (7)

Of (7) $\Rightarrow$ if width of slit $a$ increased then the width of the peak $l$ is decreases (8)

Of (8) $\Rightarrow$ the separation distance between two consecutive bright bonds will decreases.

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