Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

A standing wave pattern is formed when a plane electromagnetic wave from a transmitter falls on a perfect reflector at normal incidence. A receiver kept at a distance d from the reflector and along the line of normal incidence detects a minimum for a signal of frequency, ν = 100 MHz. When the frequency is decreased to ν = 99.9 MHz, the receiver records a maximum. Assuming c = 3 × 10^8 ms−1, the distance d is (A) 75 m (B) 1500 m (C) 3000 m (D) 750 m

A standing wave pattern is formed when a plane electromagnetic wave from a transmitter falls on a perfect reflector at normal incidence. A receiver kept at a distance d from the reflector and along the line of normal incidence detects a minimum for a signal of frequency, ν = 100 MHz. When the frequency is decreased to ν = 99.9 MHz, the receiver records a maximum. Assuming c = 3 × 10^8 ms−1, the distance d is
(A) 75 m (B) 1500 m (C) 3000 m (D) 750 m

Grade:12th pass

1 Answers

Darshit Solanki
13 Points
one year ago
Clearly, in between the receiver and the reflector the following condition is satisfied:
For v = 100 MHz, d = nλ/2; where is ‘n’ is some integer.   ------------------(1)
 
For v’ = 99.9 MHz, d = (n-1)λ’/2 + λ’/4 [Here, the n would be one less compared to the above case. P.S. – There is no option to upload a picture here so can’t help much with the explanation.]    -------------------------(2)
 
from (1) and (2)
 
nλ/2 = (n-1)λ’/2 + λ’/4
 
nc/v = (n-1)c/v’ + c/2v’
 
Upon simplification you will get,
 
n = v/2(v-v’) putting and (1) and simplifying, you get
 
d = c/4(v-v’) = 750m.   Ans. Option (D)
 

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free