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Refraction, diffraction and interference

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Question 7

Light from an amber-emitting semiconductor diode (LAL_{\text{A}}LA​) is incident normally on a plane diffraction grating. The peak wavelength of the emitted light is λp=595 nm\lambda_{\text{p}} = 595 \text{ nm}λp​=595 nm. The second-order maximum (n=2n=2n=2) for this wavelength occurs at a diffraction angle of 48.6∘48.6^\circ48.6∘.

1.

Determine NNN, the number of lines per metre on the grating.

[4]
2.

Suggest one possible disadvantage of using a higher-order maximum, such as the second-order maximum, to determine NNN with an LED light source.

[1]
3.

The relationship between the activation voltage VAV_{\text{A}}VA​ of the LED and its peak wavelength λp\lambda_{\text{p}}λp​ is given by:

VA=hceλp V_{\text{A}} = \frac{hc}{e\lambda_{\text{p}}} VA​=eλp​hc​

Calculate the activation voltage VAV_{\text{A}}VA​ for this LED. (Take h=6.63×10−34 J sh = 6.63 \times 10^{-34} \text{ J s}h=6.63×10−34 J s, c=3.00×108 m s−1c = 3.00 \times 10^8 \text{ m s}^{-1}c=3.00×108 m s−1, and e=1.60×10−19 Ce = 1.60 \times 10^{-19} \text{ C}e=1.60×10−19 C).

[2]
4.

The LED is connected in series with a resistor of resistance RRR across a power supply of emf 5.00 V5.00 \text{ V}5.00 V and negligible internal resistance. The current must not exceed 18.0 mA18.0 \text{ mA}18.0 mA. At this current, the potential difference across the LED is 2.15 V2.15 \text{ V}2.15 V. Determine the minimum value of RRR.

[3]

Refraction, diffraction and interference Questions

  1. A Level
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  3. /Refraction, diffraction and interference