Question

If a soap film (n = 1.33) has its longest constructive reflection at the red end of the visible spectrum (700 nm), by how much must it thin (in nm) to shift the longest constructive reflection to the blue end of the spectrum (400 nm). Note that we are asking for the difference in thickness, and answer with a positive number.

Answer 1

Answer:

75.19 nm

Explanation:

Condition of constructive interference of the light reflecting from the top of the thin film.

$2ndcos\theta=(m-\frac{1}{2} )\lambda$

m=1,2,3...

d= thickness of the film

n= refractive index of film

λ= wavelength of the light incident on the film.

for the longest central maximum interference, m=1

⇒$2ndcos\theta=(\frac{\lambda}{2} )$

⇒$d=\frac{\lambda}{4n} =\frac{700}{4\times1.33}= 131.6\text{_nm}$  nm

similarly, for the same order of reflection the thickness of the film=d'

and wavelength of constructive interference= λ'=400 nm

n=1.33

$2nd'cos\theta=(m-\frac{1}{2} )\lambda'$

⇒$d'=\frac{\lambda'}{4n}= \frac{400}{4\times1.33}$=  75.19 nm

Answer 2

Answer:

The difference in thickness is 56.32 nm.

Explanation:

Given that,

Wavelength = 700 nm

Refractive index = 1.33

The constructive interference of light reflecting from the top of thin film

Using formula for constructive interference

$2nd\cos\theta=(m-\dfrac{1}{2})\lambda$

Where, d = thickness of film

n = refractive index of film

$\lambda$= wave length of incident light

Here, m= 1,2,3....

For the longest interference, m = 1

$d=\dfrac{\lambda}{4n}$

We need to calculate the distance

Put the value into the formula of distance

$d= \dfrac{700}{4\times1.33}$

$d=131.5\ nm$

(b). Now, wavelength = 400 nm

We need to calculate the distance for same order

Again put the value into the formula of distance

$d'=\dfrac{400}{4\times1.33}$

$d'=75.18\ nm$

We need to calculate the difference in thickness

Using formula of thickness

$d''=d-d'$

$d''=131.5-75.18$

$d''=56.32\ nm$

Hence, The difference in thickness is 56.32 nm.