Answer
(a) Less than
(b) $454.5nm$
Work Step by Step
(a) We can see in the given figure that the separation between the fringes for the second color is less than the first one. Hence, the angle between the fringes for the second color is less as compared to the first. We also know that the wavelength is proportional to $sin\theta$. Therefore, the wavelength for the second color is less than the first color, $(505nm)$.
(b) We know that
$\frac{y}{L}=\frac{9\lambda_1}{2d}$....eq(1)
and $dsin\theta_2=m\lambda$
for $m=5$
$dsin\theta_2=5\lambda_2$
$\implies sin\theta_2=\frac{5\lambda_2}{d}$
As $\theta$ is small, therefore, $tan\theta_2\approx sin\theta_2$
$\implies \frac{y}{L}=\frac{5\lambda_2}{d}$....eq(2)
Comparing eq(1) and eq(2), we obtain:
$\frac{5\lambda_2}{d}=\frac{9\lambda_1}{2d}$
This simplifies to:
$\lambda_2=(\frac{9}{10})\lambda_1$
We plug in the known values to obtain:
$\lambda_2=(\frac{9}{10})(505nm)=454.5nm$
