# What is the angular width of the central maximum?

A single slit of width 1.50 um is illuminated with light of wavelenth 500.0 nm. What is the angualar width of the central maximum?

## Answers and Replies

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Hootenanny
Staff Emeritus
Gold Member
Please show some work or initial thoughts.

~H

the answer in the book is 38.9 degrees but how is this right?

I tried the formula &= wavelenth / b and came up with (5 * 10^-9) / (1.5 *10^-6) and got 3.33 * 10^-3 then multiplied this by 2. is this right?

anymore ideas anyone ???

Hootenanny
Staff Emeritus
Gold Member
The answer in the book is correct. You are simply a factor of $10^{2}$ out with respect to the $\sin\theta$. Your error lies here:

wayneo said:
(5 * 10^-9) / (1.5 *10^-6)
$$500nm \neq 5\times 10^{-9}m$$

Recheck your calculation. Also you need to find the angle (ArcSin)

~H

Last edited:
nrqed
Homework Helper
Gold Member
wayneo said:
I tried the formula &= wavelenth / b and came up with (5 * 10^-9) / (1.5 *10^-6) and got 3.33 * 10^-3 then multiplied this by 2. is this right?
first, 500 nm = 500 times 10^-9.

The second thing is that what you found is sin(theta). To find the *angle* you must do the inverse sin of that. Then multiply the result (in degrees) by two

thanks guys it works, just made a silly error!