Answer:
The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Explanation:
Given that,
Wavelength = λ
For D to be small,
We need to calculate the minimum width
Using formula of minimum width


Where, D = width of slit
= wavelength
Put the value into the formula

Here,
should be maximum.
So. maximum value of
is 1
Put the value into the formula


(b). If the minimum number is 50
Then, the width is


(c). If the minimum number is 1000
Then, the width is


Hence, The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
<h3>X-Rays contradict to?</h3>
<h3>C. gamma </h3>
a type of penetrating electromagnetic radiation produced by the radioactive disintegration of atomic nuclei
W = mg, Assuming g ≈ 9.8 m/s² on the earth surface.
735 N = m* 9.8
735/9.8 = m
75 = m
Mass , m = 75 kg. B.
Answer:
measure the vector diagram first
The correct answer that would complete the given statement above would be option B. UNIT (of measurement). <span>A standard quantity of a physical property used as a factor to compare other occurring quantities of that property is called a unit of measurement. Hope this answers your question. </span>