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:
I do not have enough information to tell
Explanation:
This is deduced due to the fact that if the net force due to B and C on A is zero, the charges on B and C could either be positive or negative depending on the charge on A.
Answer:
The right solution is "165.8 nm".
Explanation:
Given:
Index of refraction,
n = 1.81
Wavelength,
λ = 600 nm
We know that,
⇒ 
By putting the values, we get


For a lens, the following sign convention is generally used:
- f (the focal length) is positive for a converging lens and negative for a diverging lens
-

(the distance of the object from the lens) is positive if the object is in front of the lens and negative if it is behind the lens
-

(distance of the image from the lens) is positive if the image is behind the lens (real image) and negative if the image is in front of the lens (virtual image)
Therefore, the correct option is
<span>A. +di
</span>which mens that the image is real and located behind the lens.