Question 448: Wave Optics

Question

In the Young's double-slit experiment using a monochromatic light of wavelength $$\lambda$$, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is

Accepted Answer

Wavelength of light is $$\lambda$$.
Now, using the relation for intensity, we have
$$I = {I_{\max }}{\cos ^2}\left( {{\phi  \over 2}} \right)$$
$${1 \over 2} = {\cos ^2}\left( {{\phi  \over 2}} \right) \Rightarrow \cos {\phi  \over 2} = {1 \over {\sqrt 2 }}$$
$$ \Rightarrow {\phi  \over 2} = {\pi  \over 4} \Rightarrow \phi  = {\pi  \over 2}$$
$$ \Rightarrow \phi  = {\pi  \over 2},{{3\pi } \over 2},{{5\pi } \over 2},{{7\pi } \over 2}$$
$$\phi  = {{2\pi } \over \lambda }\Delta x \Rightarrow {\pi  \over 2} = {{2\pi } \over 2}\Delta x \Rightarrow \Delta x = {\lambda  \over 4},{{3\lambda } \over 4},{{5\lambda } \over 4}$$
$$ \Rightarrow \Delta x = (2n + 1){\lambda  \over 4}$$

Answer

$$(2n + 1){\lambda  \over 2}$$

Answer

$$(2n + 1){\lambda  \over 4}$$

Answer

$$(2n + 1){\lambda  \over 8}$$

Answer

$$(2n + 1){\lambda  \over {16}}$$

Wave Optics

In the Young's double-slit experiment using a monochromatic light of wavelength $\lambda$, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is

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