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## Homework Statement

A thin flake of mica (n=1.58) is used to cover one slit of a double-slit arrangement.The central point on the screen is occupied by the 7th brigth fringe.If lamda = 550nm, what is the thickness of the mica?

## Homework Equations

path difference d=(n-1)*thickness

d=7*lamda

snell's law?

## The Attempt at a Solution

first, the paths above the glass must be the same.So they are not in consideration.

let l be the thickness of the glass,t be the length travelled by light in mica,x in air

The mica:

sin

**θ=nsin**

(1)/(2),

sin

**∅**

**sin****∅=sin****θ/n....(1)****sin****∅=l/t..(2)**(1)/(2),

sin

**θ=nl/t...(3)**In the air:

sin

**θ=l/x...(4)**

**compare 4 and 3,**

i came out, t=xn.

Then i started to consider the difference of wavelength,

because the wavelength in mica is shorter than in air.

I tried to subtract the sum of the pulses in air and mica and set it equals to 7 lamda,

but what came out is peculiar,

let m , n be the number of pulses in mica and air respectively,f=frequency

m=tf/v...(5)

n=xf/c..(6)

so....

m/n=nt/x

i suppose it being m/n = t/nx ......so that they can be cancelled out......

Please help for this:((

i came out, t=xn.

Then i started to consider the difference of wavelength,

because the wavelength in mica is shorter than in air.

I tried to subtract the sum of the pulses in air and mica and set it equals to 7 lamda,

but what came out is peculiar,

let m , n be the number of pulses in mica and air respectively,f=frequency

m=tf/v...(5)

n=xf/c..(6)

so....

m/n=nt/x

i suppose it being m/n = t/nx ......so that they can be cancelled out......

Please help for this:((