1.

The pass-axes of two polarisers were kept such that the incident unpolarised beam of intensity I_0 , gets completely blocked. Another polariser was introduced in between these two polarisers with its pass-axis 60^@with respect to the pass-axis of the first one. The output intensity would then become

Answer»

`0`
`3/32 l_0`
`3/16 l_0`
`3/8 l_0`

SOLUTION : Since the pass axes of two polarisers `P_1` and `P_2`were KEPT such that the incident unpolarised beam of intensity `I_0`gets completely blocked. This means that both polarisers were kept cross (perpendicular of their pass axes) to each other. When third polariser `P_3`is introduced between `P_1` and `P_2`such that the angle between the pass axes of `P_1` and `P_3`is `60^@` , i.e.`theta_1 = 60^@`
intensity of polarised light emerging from first polaroid, `I_1 = (I_0)/(2)`
Intensity of polarised light (`I_3` ) emerging from polaroid `P_3`is given by law of Malus,
i.e.,`I_3 = I_1 cos^2 THETA`
` = (I_0)/(2) cos^2 60 = (I_0)/(2) 1/4`
`I_3 = (I_0)/(8)`
Now, the angle between pass axis of `P_3`and `P_2`
` theta_2 = 90^@ - 60^@ = 30^@`
` therefore ` Intensity of polarised light emerging from lost (second) polaroid `P_2`
`I_2 = I_3 cos^2 30^@= (I_0)/(8) . 3/4 = (3I_0)/(32)`


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