JEE Main 17 March 2021 Shift 2 Solved Paper

Intensity of the electromagnetic radiation is given as
Here,
U‌avg ‌= average energy density,
C= speed of radiation in air (or vacuum),
ε0= permittivity of the free space
and E0= peak value of the electric field.
⇒‌‌P∝E02⇒‌

P1

P2

=‌

E12

E22

⇒‌‌‌

E1

E2

=√‌

P1

P2

Substituting the values in the above equation, we get
⇒‌‌‌

E1

E2

=√‌

60

100

⇒‌‌E1=√‌

3

5

E2=√‌

3

5

=E (∵E2=E)
Comparing with √‌

x

5

E, we get
x=3