State why the light reflected from the two microscope slides produces a system of interference fringes.

The condition that a bright fringe is observed in the field of view of the travelling microscope is given by the relationship

$$2t = \left( {m + \frac{1}{2}} \right)\lambda$$

where $t$ is the thickness of the air film formed by the wedge at the point where the bright fringe is observed, $m$ is an integer and $\lambda$ is the wavelength of the incident light.

State the reason for the factor $\frac{1}{2}$ in the relationship.

In the diagram, the length of the slides is 5.00 cm. The wavelength of the monochromatic light is $5.92 \times {10^{ - 7}}{\text{ m}}$. Using the travelling microscope it is observed that 50 fringes occupy a length of 0.940 cm. Show that the diameter of the hair used to separate the slides is about ${\text{80 }}\mu {\text{m}}$.

light reflected from the top slide interferes with light reflected from the bottom slide;

the light reflected from the bottom slide undergoes a $\pi$ change in phase;

in moving from one (bright) fringe to the next the thickness of the air film changes by $\frac{\lambda }{2}$;

in 5.0 cm number of fringes $= \frac{5}{{0.940}} \times 50 = 266$;

therefore diameter of hair $= 133 \times 5.92 \times {10^{ - 7}} = 7.87 \times {10^{ - 5}}{\text{ m}}$;

$80{\text{ }}\mu {\text{m}}$

A significant number of candidates did not attempt this question or made very poor attempts at answering.

A significant number of candidates did not attempt this question or made very poor attempts at answering.

A significant number of candidates did not attempt this question or made very poor attempts at answering. Candidates who gave good answers often used, correctly, a similar triangles approach to determining the diameter of the hair in part (c).

Outline the process by which coloured fringes are formed.

The following data are available:

Refractive index of oil = 1.4
Refractive index of water = 1.3
Thickness of the oil film = 250 nm

Calculate the maximum wavelength of the incident light for which destructive interference occurs.

Explain why the film of oil appears to show coloured fringes.

When white light is normally incident on the surface of the oil, the film appears green to an observer. The wavelength of green light in air is 520 nm. Calculate the thickness of the film of oil.

the light reflects at the air:oil and the oil:water boundary; (can be shown on diagram)

light reflecting from the oil:water boundary interferes with the light reflecting from the air:oil boundary;

depending on the thickness of the oil, different frequencies show constructive/destructive interference (hence coloured fringes);

for constructive interference:

$\left( {2nt = \left[ {m + \frac{1}{2}} \right]\lambda } \right)$

$t = \left[ {m + \frac{1}{2}} \right]\frac{\lambda }{{2n}}$$\,\,\,$or$\,\,\,$$\left[ {0 + \frac{1}{2}} \right]\frac{{520 \times {{10}^{ - 9}}}}{{2 \times 1.40}}$;

$t = 93{\text{ nm}}$;

A large number of candidates did not explain two reflected rays, and that interference was dependent on the thickness of the oil film.

A large number of candidates did not explain two reflected rays, and that interference was dependent on the thickness of the oil film.

Describe the motion of the spring-mass system.

determine the initial energy.

calculate the Q at the start of the motion.

damped oscillation / OWTTE

[1 mark]

E «= $\frac{1}{2}$ × 30 × π² × 0.8²» = 95 «J»

Allow initial amplitude between 0.77 to 0.80, giving range between: 88 to 95 J.

[1 mark]

ΔE = 95 – $\frac{1}{2}$ × 30 × π² × 0.72² = 18 «J»

Q = « 2π$\frac{{95}}{{18}}$ =» 33

Accept values between 0.70 and 0.73, giving a range of ΔE between 22 and 9, giving Q between 27 and 61.

Watch for ECF from (b)(i).

[2 marks]

Determine the minimum thickness of the oil layer that gives rise to the least amount of light being reflected.

Describe the change in the intensity of the reflected light as the thickness of the oil layer in (a) is gradually increased.

use of $m = 1$;

$2 \times 1.51 \times t = 1 \times 579 \times {10^{ - 9}}$;

$t = 1.92 \times {10^{ - 7}}{\text{ (m)}}$;

Award [3] for a bald correct answer.

Award [2 max] for use of $m = \frac{1}{2}$ giving 9.6 $\times$ 10^–8 (m).

Award [2 max] for answer of $\frac{\lambda }{2}$ for air (2.9 $\times$ 10^–7 (m)).

intensity increases;

intensity then decreases and increases repeatedly;

when thickness becomes very large the intensity becomes constant;

In (a) the correct thin film formula from the data booklet was sometimes chosen, but many candidates were able to answer the question using first principles. A common mistake was was to give the half wavelength value in air rather than oil or to forget the phase change on the hard reflection.

Few could give a well reasoned answer to (b) and were often lucky to score one mark.

Calculate the wavelength of the light within the soap solution.

Calculate the minimum thickness of the soap film that results in constructive interference for the reflected light.

Without a calculation, explain why a soap film that is twice as thick as that calculated in (b) results in destructive interference.

$\lambda ' = \frac{\lambda }{{1.33}} = \frac{{572}}{{1.3}} = 440{\text{ nm}}$;

110 nm;

there would be a full wavelength within the film;

but the phase change at the first surface means that there is always destructive interference;

Vast majority of candidates calculated the values well and also explained destructive interference properly.

Vast majority of candidates calculated the values well and also explained destructive interference properly.

Vast majority of candidates calculated the values well and also explained destructive interference properly.

An observer viewing the microscope slide at near-normal incidence measures the fringe spacing to be 0.29 mm. Calculate the thickness of the hair.

there are $\frac{{60}}{{0.29}}$ fringes=207;
2×1×t=207×5.9×10^-7;
t=61(μm);

or

$\left( {\tan \theta  = } \right)\frac{1}{{6.0\left( {{\rm{cm}}} \right)}} = \frac{{0.5\lambda }}{{\Delta x}}$;
$t = \frac{{\left[ {0.06\left( {\rm{m}} \right) \times 0.5 \times 5.9 \times {{10}^{ - 7}}\left( {\rm{m}} \right)} \right]}}{{0.00029\left( {\rm{m}} \right)}}$;
t=61(μm);
A phase change of $\frac{1}{2}\lambda$, if seen in working, can be ignored and does not affect the answer.
Award [3] for a bald correct answer.

Deduce that the thickness of the air wedge t that gives rise to a bright fringe, is given by $2t = (m + \frac{1}{2})\lambda$.

The length of the air wedge, L, is 8.2 cm. The bright fringes are each separated by a distance of 1.2 mm. Calculate the thickness of the paper.

phase change of π occurs on reflection at one slide but not the other;
constructive interference occurs when path difference between two reflected rays is $\frac{\lambda }{2},\frac{{3\lambda }}{2},\frac{{5\lambda }}{2}$ etc;
the extra distance travelled is twice the thickness of the air (wedge) hence 2t = [m+½] λ;

number of fringes = $\frac{{82}}{{1.2}} = 68$;
fringe separation corresponds to a change in thickness of $\frac{\lambda }{2}$;
thickness of paper$= 68 \times \frac{{590 \times {{10}^{ - 9}}}}{2} = 2.0 \times {10^{ - 5}}{\rm{m}}$;

Outline how the fringes are formed.

State and explain how the fringe separation changes if the angle of the wedge is increased slightly.

fringes arise as a result of interference between light reflected from top and bottom plates;
there is a phase difference of $\pi$ at bottom reflection;
hence constructive interference when 2d=(n+½)λ and destructive when 2d=nλ;

the fringe separation would decrease;
the distance between the points/places/positions where the light is in phase or antiphase would be shorter;

Show that when θ=0 the condition for destructive interference between rays X and Y is

2t=(m+½)λ

where m is an integer and λ is the wavelength of light in the magnesium fluoride film.

Light of wavelength 640 nm in air is incident normally on the glass surface.

(i) Show that the wavelength of light in the magnesium fluoride film is 464 nm.

(ii) Calculate the minimum thickness of the film for which no light will be reflected back into the air.

the path difference (at normal incidence) is 2t;
since there is phase change of π at both surfaces there is destructive interference if the path difference is half integral multiple of the wavelength;
and so 2t=(m+½)λ

(i) $\lambda  = \frac{{{\lambda _{{\rm{air}}}}}}{n} = \frac{{640}}{{1.38}}$;
=464nm

(ii) $m = 0 \Rightarrow t = \frac{\lambda }{4}$;
$t = \left( {\frac{{464}}{4} = } \right)116{\rm{nm}} \approx 120{\rm{nm}}$;

State what phase change occurs on reflection at the air-coating boundary and at the coating-glass boundary.

The thickness d of the coating layer is 110 nm.

Determine the wavelength for which there is no resultant reflection from the surface of the lens for light at normal incidence (θ = 0º).

180º / π;

path difference must be $\frac{\lambda }{2}$;

physical thickness must be $\frac{\lambda }{2n}$;

so, maximum wavelength $\left( {{\rm{is}}\left[ {2nt = \left[ {m + \frac{1}{2}} \right]\lambda } \right] \to \lambda  = 4{n_c}d} \right) = 528\left( {{\rm{nm}}} \right)$;

Allow any valid alternative method.

was well done in general

Was poorly understood. There were some good answers but many candidates omitted n or put n = 1 as in a diffraction grating.

A ray of monochromatic light is incident on a thin film of soap water that is suspended in air. The diagram shows the reflection of this ray from the top and bottom surfaces of the film.

On the diagram, label, with the letter P, the point at which a phase difference of $\pi$ occurs.

White light is incident normally on the soap film. The thickness d of the soap film is 225 nm and its refractive index is 1.34.

(i) Show that the longest wavelength of light λ in air for which the reflected rays destructively interfere is 603 nm.

(ii) Explain why the soap film will appear coloured.

P as shown;

(i) for destructive interference, use of 2μd=nλ to give λ=2μd (ie n=1);
λ=2×1.34×225;
(603nm)
Answer given, look for correct working.

(ii) the reflected light is white minus the wavelengths that suffer destructive interference;
some colours are determined by the missing wavelengths;
some colours are enhanced due to constructive interference; 

There is considerable uncertainty amongst candidates about the surface at which a phase change occurs on reflection.

(b)(i) was an easy two marks for most. However (b)(ii) was poorly answered. More often than not only refraction, diffraction or dispersion were mentioned - even though the G5 question was clearly indicated as being about thin film interference. It is very good practice for candidates to read the stem of the question carefully before starting, and to highlight key phrases or data.

Calculate the wavelength of the light.

The upper glass plate is now replaced with a curved glass plate. The dotted line represents the upper glass plate used in (a).

Sketch the new fringe pattern in the space below. The fringe pattern of (a) is given for comparison.

$\lambda  = \left( {2D\tan \theta  = } \right)2 \times 0.30\tan {10^{ - 3}}$ or $2 \times 0.30\sin {10^{ - 3}}$;
$\lambda  = 6.0 \times {10^{ - 7}}\left( {\rm{m}} \right)$;
Award [1 max] for use of degrees instead of radians giving $\lambda  = 1.0 \times {10^{ - 8}}\left( {\rm{m}} \right)$.

decreasing distance from left to right;
distance larger than original at left and shorter than original at right;