Answer:
<u>Development in science:</u>
There a number of individuals in the world around us who has improved the field of science and made it easy for us to know this universe more easily then ever. This universe is still unexplored and there are going to be more then billions of stars which are still not studied and has an immense amount of information regarding the universe.
<u>Stephen Hawking:</u> The scientist who made it easy for each of us to know the universe and more over the cosmos in a more detailed form, as before this no one knew about the different patterns of entities and there properties.
<u>A genius who was handicapped:</u>
He was a genius in cosmology, mathematics, and other subjects of science as he was diagnosed by the amyotrophic lateral sclerosis(ALS), which made it unable for him to live a normal life.But, he communicated through a computer which detected his nerve signals and shared his thoughts about any thing or subject.
<u>Why him?</u>
Stephen hawking was just tremendous in making it more understandable about the black holes, time worms, and space etc.As it was known to very small number of people before he took the stage.
Answer:
a) m =1 θ = sin⁻¹ λ / d, m = 2 θ = sin⁻¹ ( λ / 2d)
, c) m = 3
Explanation:
a) In the interference phenomenon the maxima are given by the expression
d sin θ = m λ
the maximum for m = 1 is at the angle
θ = sin⁻¹ λ / d
the second maximum m = 2
θ = sin⁻¹ ( λ / 2d)
the third maximum m = 3
θ = sin⁻¹ ( λ / 3d)
the fourth maximum m = 4
θ = sin⁻¹ ( λ / 4d)
b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.
I = I₀ cos² (Ф) (sin x / x)²
Ф = π d sin θ /λ
x = pi a sin θ /λ
where a is the width of the slits
with the values of part a are introduced in the expression and we can calculate intensity of each maximum
c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present
maximum interference d sin θ = m λ
first diffraction minimum a sin θ = λ
we divide the two expressions
d / a = m
In our case
3a / a = m
m = 3
order three is no longer visible
I’m sorry I don’t understand this language
