Question

At a beach the light is generallypartially polarized owing to reflections off sand and water. At a particular beach on a particular day near sundown, the horizontalcomponent of the electric field vector is 2.3 times the vertical component. A standingsunbather puts on polarizing sunglasses; the glasses eliminate the horizontal field component.(a) What fraction of the light intensity received before the glasses were put on now reaches the sunbather’s eyes? (b) The sunbather, still wearing the glasses, lies on his side. Whatfraction of the light intensity receivedbefore the glasses were put on now reaches his eyes?

Answer 1

Answer:

a) 0.159

b) 0.84

Explanation:

The Horizontal component is 2.3 times the vertical component

Let the horizontal electric field component = $E_{h}$

Let the vertical electric field component = $E_{v}$

The formula for light intensity is given by:

$I = \frac{E_{m} ^{2} }{2c \mu}$..............................(1)

$E_{m}$ is the resolution of the vertical and horizontal components, $E_{h} and E_{v}$

$E_{m} ^{2} = E_{h} ^{2} + E_{v} ^{2}$..................(2)

Light intensity before the glasses were put on:

$I_{1} = \frac{E_{m} ^{2} }{2c \mu_{1} }$.............................(3)

Put equation (2) into equation (3)

$I_{1} = \frac{E_{h} ^{2} + E_{v} ^{2}}{2c \mu_{1} }$.............................(4)

After the glasses were put on the horizontal component vanishes, i.e. $E_{h} = 0$

$I_{2} = \frac{ E_{v} ^{2}}{2c \mu_{2} }$...................................(5)

Divide equation (5) by equation (4)

$\frac{I_{2} }{I_{1} } = \frac{E_{v} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$...............................(6)

But $E_{h} = 2.3E_{v}$......................(7)

Insert equation (7) into (6)

$\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{(2.3E_{v})^{2} + E_{v} ^{2} } \\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{5.29E_{v}^{2} + E_{v} ^{2} }\\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } =\frac{1}{6.29}$

$\frac{I_{2} }{I_{1} }= 0.159$

b) When the sunbather lies on his side, the vertical component vanishes, i.e $E_{v} = 0$

$\frac{I_{2} }{I_{1} } = \frac{E_{h} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$

$\frac{I_{2} }{I_{1} } = \frac{(2.3E_{v} )^{2} }{E_{v} ^{2} +(2.3E_{v} )^{2}}$

$\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2} }{E_{v} ^{2} +5.29E_{v}^{2} }$

$\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } = \frac{5.29}{6.29} \\\frac{I_{2} }{I_{1} } = 0.84$