91Ó°ÊÓ

IDENTIFY and SET UP:

This problem uses the ideas about circular diffraction In part (a) the angular radius 01 of the First dark ring in the diffraction

pattern follOWS from the geometrical relationship, where 1' is the radius of the ?rst dark ring (see Fig. 3625).

Then We use Eq- (36.17) to solve for the diameter D of the circular hole and thus calculate the average of its four values (our target variable)- In part (b) the angular rad? of the second dark ring and the third dark ring are given by the Mo

equations(Eq- 36.18), respectively. Then we use the relationship

to the radii of the second (m = 2) and third (77?. = 3) dark rings in the diffraction pattern (the target variables)-

EXECUTE:

Solving the equation : for 01 gives

But we know , then we have

Substituting the known value of A and the values of m and 7' from the given table, we ?nd

Hence, the average of the four values of D is

Therefore the average value of D is

(b) Solving the equations or 02 and 03, and respectively, and substituting the knOWn

values of A and D, We find

Hence, the radii of the second and third dark rings in the diffraction pattern are

EVALUATE:

Studying the diffraction pattern formed by using a very small hole and monochromatic light with a known wavelength

constitutes a practical way to determine the diameter of the hole-

Therefore the radii of second and third ring are 11cm and 15 cm respectively