Question

(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength $\lambda$? (b) What is its minimum width if it produces 50 minima? (c) 1000 minima?

- $D = \lambda$
- $D = 50 \lambda$
- $D = 1000 \lambda$

Solution Video

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Video Transcript

This is College Physics Answers with Shaun Dychko. We want to know the minimum width of a slit in multiples of the wavelength

*λ*that will produce a first minimum for a wavelength of*λ*. So here's the formula relating the slit width and the angle to the minimum and the order and the wavelength. So we can solve for the slit width by dividing both sides by*sin Θ*and we get then that the slit width*D*equals the order times the wavelength divided by*sin*of the angle to the minimum. So to minimize*D*, we have to maximize the denominator of this fraction the bigger the bottom of the fraction gets, the smaller the quotient becomes.*m*is prescribed to be 1 because it's the first minimum we are told in the question and*λ*is not a number, it's just going to be in terms of*λ*... it's going to be our answer. So we are going to minimize*D*by maximizing*sin Θ*now the maximum possible value for*sin Θ*is 1 because the*sin*graph looks like this and the range for the*sin*function is from negative 1 up to 1 so this is a graph of*y*equals*sin Θ*or this is the*Θ*-axis and this is the*y*-axis and you can never get above 1 so the maximum possible value is 1. So we plug in 1 for*sin Θ*, 1 for*m*and we get that*D*then is the wavelength. Now in case (b), what is the minimum width if it produces 50 minima? Well in this case,*m*is now 50 so whereas before*m*was 1 now*m*is 50 and*sin Θ*is still going to be 1 and so the answer is the slit width would have to be 50 times the wavelength. And for part (c), you'll get a 1000 minima if the slit width is 1000 times the wavelength.