Question

IV. A) An object is placed in front of a concave (diverging) lens, with del > If). Draw a ray-tracing diagram and use it to determine if the image is real/virtual and upright/inverted Divergy as Unpolanized B) -Heritelly polarized light of intensity le reaches a collection of polarizing sheets aligned as follows: Sheet 1 30° from the vertical, Sheet 2 40* from vertical, Sheet 3 horizontal. What is the intensity of the light that emerges from Sheet 3? C) A beam of light is travelling from glass (n = 1.5) to water (n = 1.33) as shown. What angle does the beam of light make with respect to the horizontal in the water? If the beam doesn't go into the water, tell me that explicitly and prove it.

Answer 1

a)We are given that the object is placed away from the focal length. Let us draw the figure.

F

Here we can see that the image formed is indeed a virtual one.

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b) Here we have to follow Malus' law to determine the intensity after every polarising sheet, ie

$I=I_0\cos^2{\theta}$

Where theta is the angle made between the previous polarsing orientation and the next.

Lets say after the light passes through sheet one the intensity will be $I_1$ and after sheet 2, it will be $I_2$ and subsequently after sheet 3 it will be $I_3$ . When an unpolarised source of light falls on a polarising sheet only half of that intensity will be transmitted since only one polarisation is allowed.

therefore, $I_1=I_0/2$

Now,

$I_2=I_1\cos^2{\theta}=\frac{I_0}{2}\cos^2({40-30})=0.48I_0$

Similarly,

$I_3=I_2\cos^2{\theta}=0.48I_0\cos^2({90-40})=0.198I_0$

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c)Given n_g=1.5 and n_w=1.33, the angle at which the beam will fail to refract into the medium will be given by Snell's law

ie,

$\frac{\sin{\theta_i}}{\sin90^{\circ}}=\frac{n_w}{n_g}$

Then,

$\theta=\sin^{-1}\left(\frac{n_w}{n_g}\right)=\sin^{-1}\left(\frac{1.33}{1.5}\right)=62.37^{\circ}$