Answer
See the detailed answer below.
Work Step by Step
For the longest wavelength, $\lambda_{max}=700$ nm, we will solve for the maximum angle of $90^\circ$.
$$d\sin\theta= m\lambda_{max}$$
Solving for $m$;
$$m=\dfrac{d\sin\theta}{\lambda_{max}}$$
Recall that the slit width $d$ is given by $1/7400$ cm,
$$m=\dfrac{\dfrac{1}{7400}\times10^{-2}\cdot \sin90^\circ}{ 700\times 10^{-9}}=\bf 1.93$$
Since $m$ must be an integer number, thus $m=1$, which means that we can see only one full order, the red light, while the second order is almost fully visible as well.
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For the smallest wavelength, $\lambda_{min}=400$ nm, we will solve for the maximum angle of $90^\circ$.
$$d\sin\theta= m\lambda_{min}$$
Solving for $m$;
$$m=\dfrac{d\sin\theta}{\lambda_{min}}$$
Recall that the slit width $d$ is given by $1/7400$ cm,
$$m=\dfrac{\dfrac{1}{7400}\times10^{-2}\cdot \sin90^\circ}{ 400\times 10^{-9}}=\bf 3.38$$
Since $m$ must be an integer number, thus $m=3$, which means that we can see part of the third order.