Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
Answer: Velocity can best be described as, the speed in a given direction.
Explanation: To find the answer, we need to know more about the Velocity of a body.
<h3>What is Velocity of a body?</h3>
- Velocity is the rate of change of displacement.
- It's a vector quantity and is measured in m/s.
- It can be positive, negative or zero.
- A body is said to be in uniform motion, then its velocity remains constant.
- Change in velocity can be a change in speed.
- The magnitude of velocity is less than or equal to speed.
Thus, we can conclude that, the option C is best describing velocity.
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Answer:
0.657 seconds
Explanation:
speed of wave= wavelength / time period
so
time period= wavelength / speed
= 4.6/7
=0.657 sec
The acceleration of the particle at 3s is [tex]a = 6 \beta [/tex]
<h3>How to calculate acceleration </h3>
if Time is given as 3s
therefore, Acceleration is
Acceleration is
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