Answer:
option D
Explanation:
given,
wavelength = 600 nm
width of separation = 0.02 mm
L = 5 m
for mth order maxima
[tex]d \times \dfrac{y_m}{L}=m\lambda[/tex]
for (m+1)th order maxima
[tex]d \times \dfrac{y_{m+1}}{L}=(m+1)\lambda[/tex]
now,
[tex]y_m=\dfrac{mL\lambda}{d}[/tex] and
[tex]y_{m+1}=\dfrac{(m+1)L\lambda}{d}[/tex]
hence,
[tex]\Delta y = y_{m+1} - y_m[/tex]
[tex]\Delta y =\dfrac{L\lambda}{d}[/tex]
[tex]\Delta y =\dfrac{5 \times 600 \times 10^{-9}}{0.02 \times 10^{-3}}[/tex]
[tex]\Delta y =0.15\ m[/tex]
[tex]\Delta y =15\ cm[/tex]
hence, the correct answer is option D