## College Physics (4th Edition)

The correct answer is: C. $6.0~m$
We can write two expressions for $L$: $L = \frac{n\lambda_n}{4} = \frac{(n)(8)}{4} = 2n$ $L = \frac{(n+2)~\lambda_{n+2}}{4} = \frac{(n+2)(4.8)}{4} = 1.2~n+2.4$ We can equate the two expressions to find $n$: $2n = 1.2n+2.4$ $0.8n = 2.4$ $n = 3$ Then: $~L = 2n = (2.0~m)(3) = 6.0~m$ The correct answer is: C. $6.0~m$