A larger population variance widens the confidence interval proportionally

X is a normal $(\mu,\sigma^2)$ random variable and a 95% confidence interval for $\mu$ is constructed based on a sample of size $n$. Suppose the length of the interval is $L_1$. If instead the variance of X is $\sigma_1^2>\sigma^2$ the length of the 95% confidence interval constructed using the same technique will be: