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How many slits per centimeter must a grating have if there are to be no second-order maxima or minima for any visible wavelength \((400 .-700 . \mathrm{nm}) ?\)

Short Answer

Expert verified
Answer: The grating must have more than \(\frac{1}{2(700 nm)}\) slits per centimeter.

Step by step solution

01

Understand the grating equation

The grating equation is given by: \[m \lambda = d \sin(\theta)\] where: - \(m\) is the order number of the maxima/minima - \(\lambda\) is the wavelength of light - \(d\) is the distance between two adjacent slits (the reciprocal of the number of slits per centimeter) - \(\theta\) is the angle at which the maxima/minima occurs
02

Set the order number and wavelength range

We want to ensure there are no second-order maxima or minima for any visible wavelength (400 nm to 700 nm). So, we set \(m = 2\), \(\lambda_{min} = 400\) nm, and \(\lambda_{max} = 700\) nm.
03

Find the angle range of the second order

To find the angle range, use the grating equation with the given wavelength range: For \(\lambda_{min}\): \[2(400 nm) = d \sin(\theta_{min})\] For \(\lambda_{max}\): \[2(700 nm) = d \sin(\theta_{max})\]
04

Ensure no second-order maxima or minima occur

To prevent any second-order maxima or minima from occurring, we must make sure the angle range for second order lies outside the possible angle range (0° to 90° or 0 to \(\pi/2\) radians). That is, \(\theta_{max} > \pi/2\) radians.
05

Calculate distance between slits using the grating equation

Using the grating equation for \(\lambda_{max}\) and the condition \(\theta_{max} > \pi/2\) radians: \[d \sin(\theta_{max}) > 2(700 nm)\] Since we are assuming \(\theta_{max}> \pi/2\), \(\sin(\theta_{max}) > 1\). Then: \[d > \frac{2(700nm)}{1}\]
06

Determine the number of slits per centimeter

The number of slits per centimeter is the reciprocal of the distance between slits: \[\frac{1}{d} < \frac{1}{2(700 nm)}\] Therefore, the grating must have more than \(\frac{1}{2(700 nm)}\) slits per centimeter to prevent any second-order maxima or minima for any visible wavelength from 400 nm to 700 nm.

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Most popular questions from this chapter

The thermal stability of a Michelson interferometer can be improved by submerging it in water. Consider an interferometer that is submerged in water, measuring light from a monochromatic source that is in air. If the movable mirror moves a distance \(d=0.200 \mathrm{~mm},\) exactly \(N=800\) fringes are shifted on the screen. What is the original wavelength (in air) of the monochromatic light?

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