Light from a blue semiconductor diode (LBL_{\text{B}}LB) is incident normally on a plane diffraction grating. The peak wavelength of the emitted light is λp=465 nm\lambda_{\text{p}} = 465 \text{ nm}λp=465 nm. The third-order maximum (n=3n=3n=3) for this wavelength occurs at a diffraction angle of 51.2∘51.2^\circ51.2∘.
Determine NNN, the number of lines per metre on the grating.
Suggest one possible disadvantage of using a higher-order maximum, such as the third-order maximum, to determine NNN with an LED light source.
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=eλphcCalculate 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).
The LED is connected in series with a resistor of resistance RRR across a power supply of emf 6.00 V6.00 \text{ V}6.00 V and negligible internal resistance. The current must not exceed 15.0 mA15.0 \text{ mA}15.0 mA. At this current, the potential difference across the LED is 2.74 V2.74 \text{ V}2.74 V. Determine the minimum value of RRR.
37 exam-style questions on AQA A Level Physics Refraction, diffraction and interference, covering Interference, Diffraction, and Refraction at a plane surface. Each one has a worked solution and a mark scheme showing where the marks go.