According to the question, the object is placed at 2F
The ray diagram is shown in the figure attached.
According to the figure:
Object AB is at 2F₁
First, we draw a ray parallel to principal axis.
So, it passes through focus after refraction.
We draw another ray which passes through optical center.
So, the ray will go through without any deviation.
Where both refracted rays meet is point A' and the image formed is A'B'
This image is formed at 2F₂
We can say that:
- Image is real.
- Image is inverted.
- Image is exactly the same size as the object.
Explanation:
the unit of area is called a derived unit because it is made of two fundamental unit metre and metre.
Answer:
14,300 lines per cm
Explanation:
Answer:
14,300 cm per line
Explanation:
λ400 nm to 400nm
We can find the maximum number of lines per centimeter, which is reciprocal of the least distance separating two adjacent slits, using the following equation.
mλ = dsin (θ)
In this equation,
m is the order of diffraction.
λ is the wavelength of the incident light.
d is the distance separating the centers of the two slits.
θ is the angle at which the mth order would diffract.
To find the least separation that allows the observation of one complete order of spectrum of the visible region, we use the maximum wavelength of the visible region is 700 nm.
d = mλ / sin (θ)
As we want the distance d to be the smallest then sin (θ) must be the greatest, and the greatest value of the sin (θ) is 1. For that we also use the longest wavelength because using the smallest wavelength, the longest wavelength would not be diffracted.
d = mλ / sin (θ)
d = 1 x 700nm / 1
= 700 nm
So, the least separation that would allow for the possibility of observing complete first order of the visible region spectra is 700 nm, and knowing the least separation we can find the maximum number of lines per cm, which is the reciprocal of the number of lines per cm.
n = 1/d
= 1 / 700 x 
= 1, 430,000 lines per m
= 14,300 lines per cm
<u>The maximum number of lines per cm, that would allow for the observation of the complete first order visible spectra.</u>
The minimum separation the plates can have is 6729 without ionizing the aje
Answer:
Current, I = 5.3 A
Explanation:
It is given that,
Radius of circular loop, r = 3 mm = 0.003 m
Magnetic field strength, B = 1.1 mT = 0.0011 T
We need to find the magnitude of the current that flows through the wire. The magnetic field for a current carrying wire is given by :
I = 5.25 A
or
I = 5.3 A
So, the magnitude of the current that flows through the wire is 5.3 A. Hence, this is the required solution.
