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Question 19M.2.HL.TZ2.8cii

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Date May 2019 Marks available [Maximum mark: 2] Reference code 19M.2.HL.TZ2.8cii
Level HL Paper 2 Time zone TZ2
Command term Show that Question number cii Adapted from N/A
cii.
[Maximum mark: 2]
19M.2.HL.TZ2.8cii

Monochromatic coherent light is incident on two parallel slits of negligible width a distance d apart. A screen is placed a distance D from the slits. Point M is directly opposite the midpoint of the slits.

Initially the lower slit is covered and the intensity of light at M due to the upper slit alone is 22 W m-2. The lower slit is now uncovered.

The width of each slit is increased to 0.030 mm. D, d and λ remain the same.

(cii)

Show that, due to single slit diffraction, the intensity at a point on the screen a distance of 28 mm from M is zero.

[2]

Markscheme

ALTERNATIVE 1

the angular position of this point is  θ = 28 × 10 − 3 1.5 = 0.01867 «rad»  ✔

which coincides with the first minimum of the diffraction envelope

θ = λ b = 560 × 10 − 9 0.030 × 10 − 3 = 0.01867 «rad» ✔

«so intensity will be zero»

 

ALTERNATIVE 2

the first minimum of the diffraction envelope is at  θ = λ b = 560 × 10 − 9 0.030 × 10 − 3 = 0.01867 «rad»   ✔

distance on screen is  y = 1.50 × 0.01867 = 28 «mm»  ✔

«so intensity will be zero»

 

Examiners report

Credit was often gained here for a calculation of an angle for alternative 2 in the markscheme but often the final substitution 1.50 was omitted to score the second mark. Both marks could be gained if the calculation was done in one step. Incorrect answers often included complicated calculations in an attempt to calculate an integer value.

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