# Optics hw

1. Mar 23, 2007

### Qyzren

Some homework q's i don't quite understand. Any help is appreciated.

A double slit arrangement produces fringes on a distant screen. Using the light from a sodium spectral lamp (lamda = 589 nm) the fringes have an angular separation of 0.015 degrees.
a) What is the separation of the slits?
b) What would the new angular separation be if
i) the slit separation were doubled?
ii) the distance to the screen were doubled?
iii) the entire arrangement were immersed in water (n=1.33)?
iv) the sodium lamp were replaced by an argon ion laser (lamda = 488 nm)?

2. Mar 23, 2007

### siddharth

How have you attempted to solve this question?

You need to show some work before you get help.

3. Mar 23, 2007

### Qyzren

i'm not sure if i'm doing it correctly, thus the reason why i posted.
let L = lamda so i don't have to figure out how to type lamda on this.

part a) d = mL/sin @
assume m = 1?
L=589*10^-9
d=2.25*10^-3 metres (using calc in degree mode)
d=2.25mm

bi)double the seperation means d = 4.50mm
@ = arcsin (mL/d) (keeping calc in degree mode)
@ = 0.0075 degrees

bii)i think this has something to do with the equation y = R mL/d but am unsure. (where y is the distance of the fringes on the screen and R is the distance from the slit to the screen.)

biii)putting the apparatus in water shouldn't affect it i don't think.

biv)use @ = arcsin (mL/d) where m = 1, L = 488*10^-9 metres and d 2.25*10^-3 metres gives @ = 0.012 degrees

4. Mar 23, 2007

### siddharth

Do you realize why to assume m=1? The value of theta is the angular separation for m=1 (ie, the first maximum)

Didn't check the numbers, but your idea is right. (With m=1)

If the distance to the screen is doubled, would theta change? Hint: $$\tan \theta = y/R$$

No, not right. What property of light, relevant to interference, changes when it travels in water?

Yes.

Last edited: Mar 23, 2007
5. Mar 23, 2007

### Qyzren

Hey thanks for your help. just some things i want to verify.