Question

In procedure 2: suppose red light passes through a double slit and falls on a screen. In the diffraction pattern, the distance from the central maximum to the first maximum is 5 mm.

a) The distance from the first minimum (dark spot) to the second minimum in the diffraction pattern is

between 7.5 mm and 10 mm

less than 2.5 mm    

more than 10 mm

between 2.5 mm and 5 mm

exactly 2.5 mm

between 5 mm and 7.5 mm

exactly 5 mm

exactly 7.5 mm

exactly 10 mm

b) Blue light has a higher frequency than red light. If you switch the light falling on the double slit from red to blue

the distance between the maxima decreases, the number of maxima on the screen remains the same

the distance between the maxima increases, the number of maxima on the screen decreases    

the distance between the maxima increases, the number of maxima on the screen increases

the distance between the maxima decreases, the number of maxima on the screen decreases

the distance between the maxima remains the same, the number of maxima on the screen increases

the distance between the maxima remains the same, the number of maxima on the screen remains the same

the distance between the maxima decreases, the number of maxima on the screen increases

the distance between the maxima increases, the number of maxima on the screen remains the same

the distance between the maxima remains the same, the number of maxima on the screen decreases

c) Diffraction experiments are usually done with diffraction gratings (which have many slits) instead of a double slit. The advantage of the diffraction grating over a double slit (using the same screen distance and slit width) is

there are many more maxima generated

it eliminates the central maximum    

all the maxima are farther apart

all the maxima are closer together

all the maxima are brighter

less light goes to the central maximum, and more to the other maxima

d) For this question, there is more than one correct answer. To get this question right, you must select all the correct answers!
You can change L (the distance from the slits to the screen), d (the distance between the slits) and λ (the wavelength of the light used). Which combination would double the distance between the maxima?

double L, leave λ and d unchanged

halve d, leave L and λ unchanged

halve L, halve d, halve λ

halve L, leave d and λ unchanged

double L, double λ, leave d unchanged

double λ, leave L and d unchanged

double d, halve L, leave λ unchanged

halve d, double λ, double L

double L, double λ, double d

halve d, double λ, leave L unchanged

double L, double d, leave λ unchanged

double L, halve d, leave λ unchanged

e) Suppose you are doing a double slit diffraction experiment with monochromatic (single frequency) light. Suddenly the atmosphere changes from air to a gas with a higher index of refraction. What will happen to the pattern?

the distance between the maxima will increase, the color of the light will change to a lower frequency

the distance between the maxima will decrease, the color of the light will remain the same    

the distance between the maxima will remain the same, the color of the light will change to a higher frequency

the distance between the maxima will increase, the color of the light will change to a higher frequency

the distance between the maxima will increase, the color of the light will remain the same

the distance between the maxima will remain the same, the color of the light will change to a lower frequency

the distance between the maxima will decrease, the color of the light will change to a lower frequency

the distance between the maxima will remain the same, the color of the light will remain the same

the distance between the maxima will decrease, the color of the light will change to a higher frequency

Answer 1

The distance between central maximum to the other maximums is defined as

and the distance between central maximum to the minimums is defined as

a) Since the distance from central maximun to the fist maximum is . then from equation (1)

Now, as we want to estimate the distance from central maximum to the first minimum , we consider equation (2)

where, by expression (3)

b) if the frequency of the blue light is higher to the red one   then, according to relationship

the length of the wave of both lights are

the length of the wave of the red light is higher than the the blue light. hence,

the distance between the maxima increases and the number of maxima increases to

c)

d) the distance between the maxima is

if and the other magnitudes don't change then

then, the distance between the maxima is double, too

Making a similar analysis, we can determine that the distance between the maxima is double when

• halve d, leave L and λ unchanged
• double λ, leave L and d unchanged
• double L, double λ, double d

e) the length of wave from the monochromatic light changes by the refractive index but its frequency doesn't change.

Thus, the length of the light in the gas is less than in the air

and, like in part b

it can watch that the distance between the maxima in the gas is less that the air. so, since frequency of the light does'n change due to the refractive index we could say

"the distance between the maxima will decrease, the color of the light will remain the same"