Unpolarized light with intensity S is incident on a series of polarizing sheets. The first sheet has its transmission axis oriented at 0°. A second polarizer has its transmission axis oriented at 45° and a third polarizer oriented with its axis at 90°. Determine the fraction of light intensity exiting the third sheet with and without the second sheet present.

Respuesta :

Answer:

Explanation:

Given

Initial Intensity of light is S

when an un-polarized light is Passed through a Polarizer then its intensity reduced to half.

When it is passed through a second Polarizer with its transmission axis [tex]\theat =45^{\circ}[/tex]

[tex]S_1=S_0\cos ^2\theta [/tex]

here [tex]S_0=\frac{S}{2}[/tex]

[tex]S_1=\frac{S}{2}\times \frac{1}{(\sqrt{2})^2}[/tex]

[tex]S_1=\frac{S}{4}[/tex]

When it is passed through third Polarizer with its axis [tex]90^{\circ}[/tex] to first but [tex]\theta =45^{\circ}[/tex] to second thus [tex]S_2[/tex]

[tex]S_2=S_0\cos ^2\theta [/tex]

[tex]S_2=\frac{S}{4}\times \frac{1}{2}[/tex]

[tex]S_2=\frac{S}{8}[/tex]

When middle sheet is absent then Final Intensity will be zero                    

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