Answer:
The interference pattern from a diffraction grating will have a wide, central bright band with alternate dark and bright bands on both sides, and the interference pattern will have an equally spaced dark and bright band.
Explanation:
In diffraction pattern we know that the distance between two consecutive minimum position is maximum
So here position of minimum is given as

now we have central maximum is of maximum width while all other have width of decreasing order.
While when we use Young's double slit pattern then we can say that position of all maximum and minimum intensity on the screen will be at same distance.
so here we have

so all the maximum and minimum intensity will have equal width.
Yes, most definitely.
Vine Supremacy
Answer:
The index of refraction of the first medium must be higher than the index of refraction of the second medium
Explanation:
Snell's law describes the behaviour of light at the boundary between two mediums:

where
n1 and n2 are the index of refraction of the two mediums
are the angle between the direction of the light ray and the normal to the interface
We can rewrite the condition as:

Let's assume now that the light is travelling in the first medium with a very large angle with respect to the normal to the surface, i.e.
, so that
. In this case, we have

We notice that if
, the ratio on the right is larger than 1, and so the term
should be also larger than 1: but this is not possible of course, since the sine function is always less than 1. Therefore, in this case total internal reflection occurs, because no refracted ray is produced.
Answer:
a. 
b. 25m/s
Explanation:
Let t be the time.
The velocity in the x direction that is subjected to acceleration of 4 m/s2
m/s
The velocity in the y direction
m/s
The total velocity at any time, which is the combination of both x and y vectors of velocity

b) at t = 5

Answer:
f = 293.38 Hz
1 st overtone = 586.76 Hz
2 nd overtone = 880.14 Hz
Explanation:
Mass of the string, m =3.4 g = 0.0034 kg
Length of the string, l = 90 cm = 0.90 m
Distance from the bridge, L = 62 cm = 0.62 m
Tension in the string, T = 500 N
Fundamental frequency is given by:

First overtone = 2 × f = 2 × 293.38 =586.76 Hz
Second Overtone = 3×f = 3 × 293.38 = 880.14 Hz