Explanation
$$\textbf{Hint:}$$
For single slit diffraction if the wavelength of light is $$\lambda $$, the distance between slit and screen is D and width of the slit is d, then the width of central bright fringe is given by,
$$W=\dfrac{2\lambda D}{d}$$
For double-slit interference if the wavelength of light is $$\lambda $$, the distance between slits and screen is D and distance between slits is d, then the width of central bright fringe is given by,
$$W=\dfrac{\lambda D}{d}$$
$$\textbf{Step1:}$$ Calculate width of central maxima for diffraction.
Let wavelength of light is $$\lambda$$, width of slit is $$a$$.If distance between slits and screen is $$D$$, then width of maxima in interference is
$${{W}_{1}} =\dfrac{2\lambda D}{ a}$$
$$\textbf{Step 2:}$$ Calculate the width of maxima for interference.
Let wavelength of light is $$\lambda$$, width of slit is $$a$$ then distance between slits is $$6.1 \times a$$.If the distance between slits and screen is $$D$$, then the width of maxima in interference is
$${{W}_{2}} =\dfrac{\lambda D}{6.1\times a}$$
$$\textbf{Step 3}$$ Find the number of maxima of double-slit interference that will lie within the central maxima of the single-slit diffraction.
Let us assume that the width of $$n$$ number of maxima of double-slit interference is equal to the width of central maxima of the single-slit diffraction pattern.
Then,
$$n= \dfrac{{{W}_{1}}}{{{W}_{2}}}$$
$$\Rightarrow n=\dfrac{\dfrac{2\lambda D}{a}}{\dfrac{\lambda D}{6.1\times a}}$$
$$\Rightarrow n=12.2$$
Since $$n$$ must be a whole number,
Therefore, $$n=12$$.
Thus, the number of maxima of double-slit interference that will be observed within the width of central maxima of single-slit diffraction is $$n=12$$.
$$\textbf{HINT:}$$ the measurable amount of a property, such as force, brightness, or a magnetic field.
$$\textbf{Step1:}$$width of silt is directly proptiontal to intensity of light
Width of slit $$(\mathrm{W}) \propto$$ Light intensity (I),$$\dfrac{\mathrm{W}_{1}}{\mathrm{~W}_{2}}=\dfrac{\mathrm{I}_{1}}{\mathrm{I}_{2}}=\dfrac{4}{9}=$$ $$\dfrac{4 \mathrm{k}}{9 \mathrm{k}} \quad \ldots(\mathrm{k}$$ is a constant $$)$$
$$I_{\max }=4 \mathrm{k}+9 \mathrm{k}+2 \sqrt{4 \mathrm{k} 9 \mathrm{k}}=25 \mathrm{k} \\$$
$$\mathrm{I}_{\min }=4 \mathrm{k}+9 \mathrm{k}-2 \sqrt{4 \mathrm{k} 9 \mathrm{k}}=\mathrm{k} \\$$
$$\textbf{Step 2:}$$ Dividing the expression,
$$\dfrac{\mathrm{I}_{\max }}{\mathrm{I}_{\min }}=25:1$$
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