Answer:
Width of the slit will be equal to 1.47 mm
Explanation:
We have given wavelength of the light 
Distance D = 8 m
Distance between first minimum dark fringe and the central maximum is 2 mm
So 
We have to find the width of the slit
For the first order wavelength is equal to 
, here a width of slit
So 
So width of the slit will be equal to 1.47 mm
Which excerpt are you talking about?
<span>The line that is drawn perpendicular to the point at which a wave intersects a boundary is know as the Normal .
When the normal is drawn, the incident ray makes an angle with it known as the angle of incidence and the reflected ray makes an angle with it known as the angle of incidence. These angles are always equal.
The refracted ray makes an angle with the normal known as angle of refraction. The sin of angle of incidence to the sin of angle of refraction is called the refractive index( </span>μ= <span>sin i / sin r) .
hope all of it helps you!</span>
Answer:
Yes such a frame exists: a free-fall (free-float frame) frame. This frame of reference is subject only to gravity and no forces such as electromagnetic forces or nuclear forces.
1.
m = mass of Mr. Ure = 65 kg
g = acceleration due to gravity = 9.8 m/s²
force of earth's gravity on Mr. Ure is given as
F = mg
F = 65 x 9.8
F = 637 N
2.
F = force of gravity on car = 3050 N
m = mass of the car = ?
g = acceleration due to gravity = 9.8 m/s²
force of gravity on car is given as
F = mg
3050 = m (9.8)
m = 3050/9.8
m = 311.22 kg
3.
m = mass of Mr. Rees = 90 kg
g = acceleration due to gravity = 9.8 m/s²
force of earth's gravity on Mr. Rees is given as
F = mg
F = 90 x 9.8
F = 882 N
