Intensity of a laser through a converging lens

AI Thread Summary
A laser beam passing through a converging lens with a focal length of 19 cm requires determining the distance at which its intensity increases by a factor of 6. The intensity is inversely proportional to the square of the distance from the lens. The initial intensity (I1) at the lens can be expressed as I1 = P/(π * 0.19^2), while the intensity at a distance x from the lens is I2 = P/(π * (0.19 - x)^2). Setting I2 equal to 6I1 allows for solving the distance x, which was initially miscalculated. The correct approach clarified the relationship between the intensities and the distances involved.
k3N70n
Messages
66
Reaction score
0

Homework Statement



A laser beam passes through a converging lens with a focal length of 19cm. At what distance past the lens has the laser beam's intensity increased by a factor of 6?

Homework Equations



I=\frac{c\epsilon_0 E_0^2}{2}??

The Attempt at a Solution



To be honest, I'm not even sure how to start on this question. I think I just need a hint to get started. Any help would be greatly appreciated.
 
Physics news on Phys.org
Intensity is inversely proportional to square of the distance. Assume that at focus the intensity is maximum, say Io.
What will be the intensity (I1) at the lens position?
What will be the intensity (I2) at a distance x from the focus?
What is the relation between I1 and I2?
 
rl.bhat said:
Intensity is inversely proportional to square of the distance. Assume that at focus the intensity is maximum, say Io.
What will be the intensity (I1) at the lens position?
What will be the intensity (I2) at a distance x from the focus?
What is the relation between I1 and I2?

Thanks for the quick reply.

I'm now trying to understand the formula given in my text I=\frac{P}{A}

So I'm thinking that I_1=\frac{P}{\pi r^2}?


If so then I_2=\frac{P}{\pi r^2}=6I_1=\frac{6P}{\pi 0.17^2}.
so r=0.0694m. Thus the distance from from the lense is 10.1cm. I entered this into my assignment online and was informed I was incorrect. Any idea as to where I went wrong?
 
Check your calculations. From where did get 0.17?
The equations should be I1 = P/pi*0.19^2
And I2 = P/pi(0.19- x )^2 where x is the distance from the lens. Put I2 = 6I1.Solve for x.
 
Thanks. Somehow .17 turned into .19. I've got it now. Really appreciate the help.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Replies
5
Views
1K
Replies
3
Views
1K
Replies
2
Views
1K
Replies
7
Views
3K
Replies
15
Views
2K
Replies
1
Views
1K
Replies
5
Views
923
Back
Top