Diffraction and interference patterns (HL)

Answer

At $\theta = \lambda/b$, where b is the slit width (small-angle approximation).

Answer

**Inversely** — narrower slit (smaller b) ⇒ **wider** central maximum, since $\theta = \lambda/b$.

Answer

**Directly** — longer wavelength (red) ⇒ wider spread, since $\theta = \lambda/b$.

Answer

$d\sin\theta = n\lambda$, with n = 0, 1, 2, … the order of the maximum.

Answer

Take the **reciprocal of the lines per metre**: $d = 1/(\text{lines per metre})$.

Answer

500 lines/mm = 500×10³ lines/m, so $d = 1/(500\times10^3) = 2.0\times10^{-6}$ m.

Answer

Many slits add in phase only at very precise angles, so each maximum is **narrow and bright**.

Answer

The whole number of wavelengths of path difference between **adjacent** slits ($d\sin\theta = n\lambda$).

Answer

The **grating** equation has sinθ ($d\sin\theta = n\lambda$); the single-slit small-angle result is just $\theta = \lambda/b$.

Answer

sinθ ≤ 1, so $n\lambda \le d$ — orders with $n > d/\lambda$ cannot exist.

Answer

**Twice** the first-minimum angle: a full width of about $2\lambda/b$.

Answer

Precisely **measuring wavelengths** (spectroscopy), because its sharp maxima pin down θ accurately.