In a Young’s double-slit experiment, two slits with a separation of 8.5 × 10-5 m
ID: 1479952 • Letter: I
Question
In a Young’s double-slit experiment, two slits with a separation of 8.5 × 10-5 m create an interference pattern on a screen that is 2.3 m away. If the tenth dark fringe above the central bright fringe is located 7.5 cm above the central bright fringe on the screen, what is the wavelength of the light being used? (Feel free to use the small angle approximation: sin tan.)
A) 264 nm
B) 326 nm
C) 308 nm
D) 277 nm
E) 292 nm
The answer is E) 292 nm, however I get 277 nm when I use the constructive bright interference formula. How do I find ? Thanks.
Explanation / Answer
trhe distance of the tenth dark fringe from the central bright is y10 = (10+0.5)*lamda*L/d
given that d= 8.5*10^-5 m
L = 2.3 m
lamda is the required wavelength = ?
and also y10 = 7.5 cm = 0.075 m
y10 = (10.5*Lamda*L/d)
lamda = y10*d/(10.5*L) = (0.075*8.5*10^-5)/(10.5*2.3) = 264 nm
hence the answer is A) 264 nm and not 292 nm