Test Index

JEE Main 15 Apr 2018 Shift 1 Solved Paper

© examsnet.com

Question : 21

Total: 90

A monochromatic beam of light has a frequency v=

×1012Hz and is propagating along the direction

.It is polarized along th

direction. The acceptable form for the magnetic field is :

E0
C
(
∧
i
−
∧
j
)
√2
cos[104
(
∧
i
+
∧
j
)
√2
.
→
r
−(3×1012)t]
E0
C
(
∧
i
−
∧
j
)
√2
cos[104
(
∧
i
−
∧
j
)
√2
.
→
r
−(3×1012)t]
E0
C
∧
k
cos[104
(
∧
i
+
∧
j
)
√2
.
→
r
+(3×1012)t]
E0
C
(
∧
i
+
∧
j
+
∧
k
)
√2
cos[104
(
∧
i
+
∧
j
)
√2
.
→
r
−(3×1012)t]

Go to Question:

Prev Question

Next Question