In the interference and diffraction experiment, the equation for single slit dif

Question

In the interference and diffraction experiment, the equation for single slit diffraction was given as n lambda = alpha sin theta, where lambda is the wavelength of light, alpha is the width of the slit, n is the order number, and theta is the angle position of a ___. A. bright maximum B. central maximum C. dark minimum D. banded spot In a single slit diffraction experiment, you take the following data: the distance from the slits to the screen is 1 m, the width of the central maximum is 4 cm and the widt6h of the slit is 0.03 mm. What is the wavelength of light being used in the experiment? A. 600 nm B. 6 mu m C. 6 times 10^7 m D. 600 mm In a single slit diffraction pattern. What is the relationship between the width of the central maximum and the width of the slit, a? A. the two variables are directly proportional B. there is no dependence on each other C. the central maximum is the sine of the slit width D. the two variables are inversely proportional In a double slit diffraction pattern, the interference fringes are modulated by the single slit pattern. You can calculate how many fringes are contained within the central maximum by predicting the location of a missing order, given by d/a. Assume that the distance between the slits is 0.12 mm, and the width of each slit is 0.03 mm. How many fringes do you expect within the central maximum of the diffraction pattern? A. 4 B. 7 C. 8 D. 9 The figure shows the interference

Explanation / Answer

37) A) 600 nm

given

R = 1m
W = 4 cm = 0.04 m
a = 0.03 mm = 3*10^-5 m
lamda = ?

we know, w = 2*lamda*R/a

==> lamda = w*a/(2*R)

= 0.04*3*10^-5/(2*1)

= 6*10^-7 m

= 600 nm

38) D

w = 2*lamda*R/a

w and a are inversly proportional to each other.

39) A) 4

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