Young's double slit

Question:
There is a double slit where the separation distance of slits is 0.100 mm. The distance from the slit to the screen is 200 cm. The first bright fringe is found at 1.00 cm from the central maximum. What is the wavelength of the laser beam?

Answer:
From the figure, we can redefine the values in SI units.
$d = 1.00 \times 10^{-4}$ m,
$y = 1.00 \times 10^{-2}$ m,
$L = 2.00$ m
The distance between the slit and screen, the position of the bright spot, and the angle from the slit to the spot make a right angle triangle as shown in the figure. We can find the angle $\theta$.
\[ \theta = \tan^{-1}\frac{y}{L} = \tan^{-1}\frac{1.00\times 10^{-2}}{2.00}   \\
             = 0.2865^o \]
For the bright maxima, we have the relationship:
\[ \lambda = d\sin\theta \]
where $\lambda$ is the wavelength of the laser.
Thus,
\[ \lambda = 1.00 \times 10^{-4}\sin(0.2865^o) = 5.00\times 10^{-7} \mathrm{m} \]
The results can also be expressed as 500 nm.

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