A. - (lambda) per unit length of the inner surface of the cylinder<span>b. (lambda)outer = 3 (lambda)
</span><span>c. E= 3(lambda)/(4 pi epsilon(0) r^2)</span>
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.
Answer:
Wavelength of sound in the pipe is 6.0 m
Explanation:
As we know that closed pipe is given to us so here at open end it will form antinode while at closed end it will form Node
Here we know that in closed pipes on odd harmonics will exist
so here for third harmonic we have

where L = length of the pipe
so we have

