Geometrical Optics: Solving 2 Prism Problem

[americanforest]

http://img140.imageshack.us/my.php?image=jonesf2238vg1.gif

Problem Statement: Two prisms with the same angle but different indices of refraction are put together to form a parallel sided block of glass (see the figure). The index of the first prism is n1 = 1.48 and that of the second prism is n2 = 1.72. A laser beam is normally incident on the first prism. What angle will the emerging beam make with the incident beam?

Eqn:

Lens Law,
GEOMETRYSolution:

It hits on the normal the first time so it keeps going straight down that normal after entering first index of refraction. It hits the inclined plane between the two indices at to the normal to that plane:
$$\phi=tan^{-1}(1/4)$$
and using lens' law it exits to the normal to the same plane:
$$\beta=sin^{-1}(\frac{n_1}{n_2}sin(\phi))$$
then it keeps going and hits the last one at an angle
$$\alpha=2\beta-\phi$$ (measuring clockwise from normal)
and leaves at an angle
$$\gamma=sin^{-1}(n_2sin(\alpha))$$

The geometry took forever but I got it wrong

 

[lippyka]

the first time.n1 = 1.48n2 = 1.72\phi = tan^{-1}(1/4) = 14.036\beta = sin^{-1}(\frac{n_1}{n_2}sin(\phi)) = 10.879\alpha = 2\beta-\phi = 19.798\gamma = sin^{-1}(n_2sin(\alpha)) = 20.619Therefore, the emerging beam makes an angle of 20.619 degrees with the incident beam.

 

[ogb p]

. I forgot to take into account that the beam will refract towards the normal after leaving the first prism. So the correct solution should be:

Using Snell's Law, we can calculate the angle of refraction for the first prism:
\theta_1=sin^{-1}(\frac{n_1}{n_2}sin(0))=0

Then, for the second prism, the angle of incidence is equal to the angle of refraction from the first prism:
\theta_2=\theta_1=0

Applying Snell's Law again, we can calculate the angle of refraction for the second prism:
\theta_3=sin^{-1}(\frac{n_2}{n_1}sin(0))=0

Finally, using the geometry of the prism block, we can calculate the angle of the emerging beam:
\gamma=180^{\circ}-(\theta_2+\theta_3)=180^{\circ}

Therefore, the emerging beam will be parallel to the incident beam, making an angle of 180 degrees with it. This means that the beam will exit the prism block in the same direction as it entered, without any deviation.