JEE MAIN - Physics (2020 - 4th September Morning Slot - No. 11)

For a transverse wave travelling along a straight line, the distance between two peaks (crests) is 5 m, while the distance between one crest and one trough is 1.5 m. The possible wavelengths (in m) of the are :
1, 3, 5, .....
$${1 \over 1},{1 \over 3},{1 \over 5},$$ .....
1, 2, 3, .....
$${1 \over 2},{1 \over 4},{1 \over 6},$$

Explanation

$$1.5 = \left( {2{n_1} + 1} \right){\lambda \over 2}$$ ......(1)

5 = n2$$\lambda $$ .....(2)

(1) $$ \div $$ (2)

$${{1.5} \over 5} = {{\left( {2{n_1} + 1} \right)} \over {2{n_2}}}$$

$$ \Rightarrow $$ 3n2 = 10n1 + 5

n1 = 1 ; n2 = 5 $$ \Rightarrow $$ $$\lambda $$ = 1

n1 = 4 ; n2 = 15 $$ \Rightarrow $$ $$\lambda $$ = 1/3

n1 = 7 ; n2 = 25 $$ \Rightarrow $$ $$\lambda $$ = 1/5

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