Question

FIGURE Q33.5 shows the light intensity on a viewing screen behind a single slit of width a The light's wavelength is$\lambda$. Is$\lambda <a,\lambda =a,\lambda >a$, or is it not possible to tell? Explain

Short answer

a) The light wavelength is remains the same

b)The lambda less than a is remains the same

c) The lambda equal to a is decrease

d)The lambda greater than a is increase

Step-by-step solution

The light wavelength(part a)

We will use some of the expressions from Young's Double-Slit Experiment chapter in the book in this problem.

Slits are replaced with slits, and the spacing d between adjacent slits remains constant.

Expression from the angles of bright fringes:

${\theta }_m=m\times \frac{\lambda }{d}$

For constant angular positions of the bright fringes in the interference pattern and spacing d, m represents the number of fringes that will remain constant.

Find fringe spacing(part b)

Fringe spacing expression:

Fringe spacing is independent of the number of slits, and for constant spacing d between slits, fringe spacing remains constant.

$\mathrm{\Delta }y=\frac{\lambda \times L}{d}$

 step3: Find λ=a (part c) 

We don't need a physics expression to assume that as the number of slits increases from to, the width of each fringe decreases while the spacing d remains constant.

Findλ>a(part d)

d) Intensity from expression:

$I_{max}=N^2\times I_1$

Because brightness intensity is proportional to the square of the amplitude N but not to the number of slits, the brightness of each fringe increases for constant spacing d.