If white light illuminates a diffraction grating having 790 lines/mm , over what range of angles does the visible m = 1 spectrum extend?

We are given the property of the illuminating light:

790 lines/mm

Which is also equivalent to:

790,000 lines/m

 

Taking the reciprocal of this value will give us the distance per line:

d = 1 m / 790,000 lines

d = 1.2658 * 10^-6 m 

 

Now we can solve for the range of angles using the formula
λ = d * sin θ 

sin θ  = λ / d

where λ is the wavelength of light and θ is the angle

Now we know that visible light ranges from 400 to 700 nm, therefore we use these values to find the range.

 

when λ = 400 nm = 400 * 10^-9 m

sin θ = 400 * 10^-9 m / 1.2658 * 10^-6 m 

θ = 18.42°

 

when λ = 700 nm = 700 * 10^-9 m

sin θ = 700 * 10^-9 m / 1.2658 * 10^-6 m 

θ = 33.57°

 

Therefore the range of angles is 18.42° to 33.57°.

The figure is missing but we can answer anyway.the correct answer is "releasing the string". in fact, when the person is holding the bow and the arrow, the kinetic energy of the arrow is still zero (because the arrow is not moving). the moment the arrow is released, it acquires a velocity v, so its kinetic energy will become different from zero: where m is the mass of the arrow and v its velocity.  all the other options, instead, do not increase the kinetic energy of the system: -  pulling farther back on the string : this will just increase the elastic potential energy of the system-  holding the arrow higher: this will increase the gravitational potential energy of the system (which is given by u=mgh, where h is the height above the ground)-  decreasing the mass of the arrow: this will decrease the gravitational potential energy of the system (which is given by u=mgh, where m is the mass)