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.. DATA While researching the use of laser pointers, you conduct a diffraction experiment with two thin parallel slits.

Your result is the pattern of closely spaced bright and dark fringes shown in Fig. P36.63. (Only the central portion of the pattern is shown.) You measure that the bright spots are equally spaced at centre to centre (except for the missing spots) on a screen that is from the slits. The light source was a helium–neon laser producing a wavelength of . (a) How far apart are the two slits? (b) How wide is each one?

Short Answer

Expert verified

two slits are 1.03mm apart

the value of a is 0.147mm

Step by step solution

01

Determine the distance between two slits

IDENTIFY and SET UP:

The equally spaced bright and dark fringes correspond to the double—slit interference pattern- The missing bright spots

correspond to the dark fringes of the single—slit diffraction pattern (the missing spots occur at the same angles as for the

single-slit diffraction pattern)- In part (a) We use Eq- (35.6) with m = 1 (We assume the Mo spots in the middle of the pattern

correspond to the interference orders m = 0 and i) to solve for the separation d betWeen the Mo slits (our target variable).

In part (b) We notice the ?rst missing spot (the diffraction ?rst—order dark fringe) occurs at the 7th-order bright fringe of the

double—split interference; they both have the same angular position- So we can sayWith the Mo

angles “; determined by Eq- (362) (with m = 1) and Eq- (35-4) (with m = 7), respectively. This leads to a

single equation in a single unknOWn of the width (1 of either slit (the target variable).

EXECUTE:

(a) Solving Eq- (55-6) with m = 1 for (1 gives

Substituting the knOWn values of R, A, and y1. We find

02

Determine which one is wide

(b) Using Eq (362) with m = 1 and Eq- (35.4) with m = 7, We obtain

But we argued that the two angles ; are actually equal- Solving the two equations for a and substituting the

know value for d, we find

EVALUATE:

The spacing between the dark fringes of the diffraction pattern is greater than that of the

bright (or dark) fringes of the interference pattern because.

Therefore the value of a is 0.147mm

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