When light of wavelength 700 nm is used in a single-slit diffraction pattern, the central maximum has a width of 3.5 cm. When the light source is replaced by one of wavelength 500 nm, what central width will result? Assume small angles.
A) 3.5 cm
B) 3.0 cm
C) 2.5 cm
D) 4.9 cm
E) 4.0 cm

Question

Quizplus · Accepted Answer

Using the single-slit diffraction formula: 
$$w = \frac{\lambda L}{a}$$
where $w$ is the width of the central maximum, $\lambda$ is the wavelength of light, $L$ is the distance between the slit and the screen, and $a$ is the width of the slit.

We are given that $w=3.5$ cm, $\lambda_1 = 700$ nm, and we assume $a$ and $L$ are the same. We want to find $w_2$ when $\lambda_2=500$ nm.

Using the same formula, we have:
$$w_2 = \frac{\lambda_2 L}{a} = \frac{\lambda_1}{\lambda_2}w_1 = \frac{700	ext{ nm}}{500	ext{ nm}}\cdot 3.5	ext{ cm} = \boxed{2.5	ext{ cm}}$$

When Light of Wavelength 700 Nm Is Used in a Single-Slit