JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 15)
A beam of plane polarised light of large
cross-sectional area and uniform intensity
of 3.3 Wm-2 falls normally on a
polariser (cross sectional area 3 $$ \times $$ 10-4 m2) which
rotates about its axis with an angular speed
of 31.4 rad/s. The energy of light passing through
the polariser per revolution, is close to :
cross-sectional area and uniform intensity
of 3.3 Wm-2 falls normally on a
polariser (cross sectional area 3 $$ \times $$ 10-4 m2) which
rotates about its axis with an angular speed
of 31.4 rad/s. The energy of light passing through
the polariser per revolution, is close to :
1.0 $$ \times $$ 10-5 J
1.0 $$ \times $$ 10-4 J
1.5 $$ \times $$ 10-4 J
5.0 $$ \times $$ 10-4 J
Explanation
Intensity, I = 3.3 Wmā3
Area, A = 3 Ć 10ā4 m2
Angular speed, $$\omega $$ = 31.4 rad/s
$$I = {I_0}{\cos ^2}(\omega t)$$
$$ \Rightarrow {I_{av}} = {{{I_0}} \over 2}$$
$$ \because $$ $$\left\langle {{{\cos }^2}\theta } \right\rangle $$ = $${1 \over 2}$$, in one time period
$$ \therefore $$ $$E = {{{I_0}} \over 2} \times A \times (\Delta t)$$
and $$\Delta t = {{2\pi } \over \omega } = {{2 \times 3.14} \over {31.4}} = {1 \over 5}s$$
$$ \therefore $$ $$E = {{3.3} \over 2} \times 3 \times {10^{ - 4}} \times {1 \over 5} = 1 \times {10^{ - 4}}\,J$$
Area, A = 3 Ć 10ā4 m2
Angular speed, $$\omega $$ = 31.4 rad/s
$$I = {I_0}{\cos ^2}(\omega t)$$
$$ \Rightarrow {I_{av}} = {{{I_0}} \over 2}$$
$$ \because $$ $$\left\langle {{{\cos }^2}\theta } \right\rangle $$ = $${1 \over 2}$$, in one time period
$$ \therefore $$ $$E = {{{I_0}} \over 2} \times A \times (\Delta t)$$
and $$\Delta t = {{2\pi } \over \omega } = {{2 \times 3.14} \over {31.4}} = {1 \over 5}s$$
$$ \therefore $$ $$E = {{3.3} \over 2} \times 3 \times {10^{ - 4}} \times {1 \over 5} = 1 \times {10^{ - 4}}\,J$$
Comments (0)
