Diffraction Grating - all visible light?

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To determine the smallest grating spacing for observing the entire visible spectrum (400nm to 700nm), the equation nλ = d sin θ is essential. The maximum wavelength for first-order diffraction is 700nm, and for second-order, it is 1400nm. The discussion raises questions about how to calculate the maximum angle for observing these wavelengths. Clarification on how to apply the grating equation effectively is needed. Understanding these principles will help in solving the problem accurately.
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Homework Statement


Find the smallest grating spacing that let's you see the entire visible spectrum.
400nm to 700nm comprises the visible light spectrum.
no other information is given.

Homework Equations


nλ=d sin θ

The Attempt at a Solution


I am not sure how to start as all i have been given is a spectrum of wavelengths...?
 
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Well, apparently its 700nm. And 1400 nm for the second order.
 
What is the maximum angle where a wavelength can be observed?
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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