When blue light (lambda = 500 nm) is incident on a single slit, the central brig

Question

When blue light (lambda = 500 nm) is incident on a single slit, the central bright spot has a width of 8.75 cm. If the screen is 3.21 m distant from the slit, calculate the slit width. Blue light (lambda = 500 times 10^-9 m) is shined on a single slit of width w. The central bright spot cf the resulting diffraction pattern on a screen a distance L = 3.21 m behind the single slit has a width (2y_1) = 8.75 times 10^-2 m. The relationship sin(theta) = m lambda/w gives the angular position of the dark fringes. Using trigonometry and the small angle approximation (sin(theta) tan(theta)), we can rewrite this in terms of the position of the mth dark fringe on the screen, y_m. Plugging in m = 1 and rearranging the expression for w will allow us to calculate the slit width.

Explanation / Answer

For a light of wavelength lambda, the ventral maxima's width is given by 2(lambda)D/a
0.0875 = 2*500*10^-9 *3.21/a
a = 36.685 micro m

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.