Maurice

Sorry it is a mistake the thin convex lens of the eye is a distance of 5.1 mm from the cornea and the distance from the lens cornea is 18 mm.

I use the lens equation for thin lens, it don't seems good...Maybe is it better to use the Jones vectors?

Thanks and sorry for my English.

Hello,

Can you help me?

Consider your eye as a thin convex lens of 5.1 mm and with focal length of 13 mm.

How far must be a object from your cornea to create a image on your retina?

I found 41.62 mm by using equation of thin lens 1/s0 + 1/si= 1/f, is it good?

Hello,

Can somebody help me...

A mass m is attached vertically to 2 springs, one spring at the ceiling K1 and the other one on the floor K2.

What is the normal mode angular frequency for vertical motion?

I found sqrt(5*k/m) but it don't seem good...

Thanks.

Yes,I think that the solution is that the greatest angle of incidence is 14 degrees respect to the normal.

Sorry I have some difficulties to clearly understand but it seems to be 28 degrees for the incident angle

ni.sini=n2.sinr (I=incident,r=refracted)

so 1.sini=1.5.sinr

so sini/sinr=1.5 I>r

into glass 90-r=critical angle and then?

I think it is 50 degrees

No sorry 90-50=40 degrees

Hello,
Can somebody help me?
Light enters an optical fiber(ideal glass optical fiber without cladding or buffer)at the normally cleaved end.

How calculate the greatest angle of incidence I that will result in total internal reflection of the light?

I know Snell's Law and refraction Law and I have found 21 or 42 degrees but that don't seem correct...

Thanks.

In fact there is a thin elastic string stitched to a normal string (to be considered as a continuum) and I must find the speed wave in this new thicker string.

Don't the mass of the thin elastic string, all what is found is the tension T, the linear density and also the speed wave in the normal string (v=sqrt(T/mu).

But the speed wave in the new thicker spring by using Hooke's Law F=-k.x?

At equilibrium T=-k.x, I know T and k(new length - original length),so I can find k(spring constant) but then to find the new wave speed,for me it is the same consider the same new length and the negligible mass of the thin elastic string?

Indeed I have some difficulties to understand clearly the wave equation and the solutions. One of the solutions is a sinusoid and take the first derivative to find the wave speed but I don't understand how to use the amplitude A, the angular frequency w and t?

Thanks.

In fact I found the tension, the mass density and the wave speed in the string but what I don't understand is if I add a thin elastic string and if I stitched the thin elastic string at a normal string (the 2 are considered as a continuum), how can I find the wave speed in the thin elastic string by using Hooke's Law and considering the thin string as A SPRING and without knowing the linear elasticity

First sorry for my English.

If I know the tension T in a elastic string, the initial length L , the length after stretching (L+l),the mass density, how could I find the speed wave in the string by using Hooke's Law and considering the string as a spring,

Thanks.