Chapter 24 - Physical Optics: The Wave of Nature of Light - Learning Path Questions and Exercises - Exercises - Page 840: 8

a). $4.4\times10^{-5}cm$ b). $0.0183mm$

a). L = 125cm d=0.0025cm Distance of third maximum from central maximum = $6.6cm$ n=3. Now, $y_{n}=\frac{nLv}{d}$, v being the wavelength $y_{3}=\frac{3\times 125 v}{0.0025}$ Thus, $v=\frac{6.6\times 0.0025}{3\times125}=4.4\times10^{-5}cm$ b). $y_{2}=\frac{2Lv}{d}=\frac{2\times125\times4.4\times10^{-5}}{6.6}=0.0183mm$