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Question 21M.2.HL.TZ1.6

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Date May 2021 Marks available [Maximum mark: 7] Reference code 21M.2.HL.TZ1.6
Level HL Paper 2 Time zone TZ1
Command term Calculate, Explain, State Question number 6 Adapted from N/A
6.
[Maximum mark: 7]
21M.2.HL.TZ1.6

A painting is protected behind a transparent glass sheet of refractive index nglass. A coating of thickness w is added to the glass sheet to reduce reflection. The refractive index of the coating ncoating is such that nglass > ncoating > 1.

The diagram illustrates rays normally incident on the coating. Incident angles on the diagram are drawn away from the normal for clarity.

(a)

State the phase change when a ray is reflected at B.

[1]

Markscheme

«change is» π/180° 

 

(b)

Explain the condition for w that eliminates reflection for a particular light wavelength in air λair.

[3]

Markscheme

«to eliminate reflection» destructive interference is required

phase change is the same at both boundaries / no relative phase change due to reflections

therefore 2wncoating=m+12λair

OR

w=λcoating4

 OR

w=λair4ncoating ✓

(c.i)

State the Rayleigh criterion for resolution.

[1]

Markscheme

central maximum of one diffraction pattern lies over the central/first minimum of the other diffraction pattern

(c.ii)

The painting contains a pattern of red dots with a spacing of 3 mm. Assume the wavelength of red light is 700 nm. The average diameter of the pupil of a human eye is 4 mm. Calculate the maximum possible distance at which these red dots are distinguished.

[2]

Markscheme

θ=«1.22λb=1.22700×10-94×10-3=»2.14×10-4«rad» ✓

D=14«0.05m» ✓