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This is one of our many quizzes on Diffraction. Click on this to access more such competitive quizzes on Wave Optics.

Diffraction Quiz 2

physics

 The corresponding frequency of light used for its diffraction response to have the product of gap width and sine of angular spread of fourth secondary maximum as $3.25\times {{10}^{-10}}\text{ m}$ is  

 Recall the condition for ${{n}^{th}}$ secondary maximum in diffraction through a single slit and proceed. 

  For ${{n}^{th}}$ secondary maximum, product of gap width and sine of angular spread is given by $\\a\sin {{\theta }_{n}}=(2n+1)\dfrac{\lambda }{2}\\ $ Therefore, for fourth secondary maximum, product of gap width and sine of angular spread is given by $\\a\sin {{\theta }_{4}}=(2\times 4+1)\dfrac{\lambda }{2} \\\Rightarrow a\sin {{\theta }_{4}}=\dfrac{9\lambda }{2} \\ $ Clearly, frequency of light from its wavelength can be derived as: $\\\dfrac{9\lambda }{2}=3.25\times {{10}^{-10}} \\\Rightarrow \lambda =\dfrac{6.5\times {{10}^{-10}}}{9} \\\Rightarrow \lambda =0.72\times {{10}^{-10}}\text{ m} \\\Rightarrow \dfrac{c}{\nu }=0.72\times {{10}^{-10}}\text{ m}\left( \because \lambda \nu =c \right) \\\Rightarrow \nu =\dfrac{3\times {{10}^{8}}}{0.72\times {{10}^{-10}}} \\\therefore \nu =4.16\times {{10}^{18}}\text{ Hz}$

 Suppose that a wave of $5252\times {{10}^{-10}}\text{ m}$ undergoes diffraction from an aperture on a metal plate. What is the size of the aperture if the reciprocal of angular deviation of the third secondary minimum is $270\text{ ra}{{\text{d}}^{-1}}$

Substitute the given values of required parameters in the expression for ${{n}^{th}}$ secondary minimum in the case of diffraction from single slit and proceed. 

  Wavelength of light used is given by $\\\lambda =5252\times {{10}^{-10}}\text{ m}\\ $ Angular deviation of third secondary minimum from its reciprocal is given by $\\\dfrac{1}{{{\theta }_{3}}}=270\text{ ra}{{\text{d}}^{-1}} \\\Rightarrow {{\theta }_{3}}=\dfrac{1}{270} \\\Rightarrow {{\theta }_{3}}=3.7\times {{10}^{-3}}\text{ rad} \\ $ Now, for ${{n}^{th}}$ secondary minimum, $\\a\sin {{\theta }_{n}}=n\lambda\\ $ Therefore, for third secondary minimum, $\\a\sin {{\theta }_{3}}=3\lambda  \\\Rightarrow a{{\theta }_{3}}=3\lambda \text{(}\because \text{sin}{{\theta }_{3}}\approx {{\theta }_{3}}) \\\Rightarrow a=\dfrac{3\lambda \text{ }}{{{\theta }_{3}}} \\\Rightarrow a=\dfrac{3\times 5252\times {{10}^{-10}}}{3.7\times {{10}^{-3}}} \\\therefore a=4.2\times {{10}^{-4}}\text{ m}$

 A lens is attached beside an opening of size $0.15\text{ cm}$. If $3200\times {{10}^{-10}}\text{ m}$ is passed through this opening, what is the focal length of the lens? Given that the distance between central maximum and first secondary minimum is $0.3\times {{10}^{-6}}\text{ m}$. 

 Use the expression for width of central maximum in diffraction by treating screen distance as focal length of the lens. 

 Size of the opening is given by $\\a=0.15\times {{10}^{-2}}\text{m}\\ $ Wavelength used is given by $\\\lambda =3200\times {{10}^{-10}}\text{m}\\ $ In diffraction, width of central maximum from distance between central maximum and first secondary minimum can be derived as:$\\x=0.3\times {{10}^{-6}}\text{ m} \\\Rightarrow 2x=0.6\times {{10}^{-6}}\text{ m} \\ $ Width of central maximum is also given by $\\2x=\dfrac{2f\lambda }{a} \\\Rightarrow f=\dfrac{2xa}{2\lambda } \\\Rightarrow f=\dfrac{0.6\times 10{}^{-6}\times 0.15\times {{10}^{-2}}}{6400\times {{10}^{-10}}} \\\therefore  f=1.4\times {{10}^{-3}}\text{ m}$

 Diffraction of a beam of a certain frequency of light becomes observable after $2.2\text{ m}$. If the size of the tear in an opaque polymer through which this frequency is passed is ${{10}^{-2}}\text{ m}$, determine this frequency. 

 Recall the concept of Fresnel distance to arrive at the required frequency of light used. 

 Distance after which diffraction becomes observable is known as Fresnel distance. It is given by $\\Z=\dfrac{{{a}^{2}}}{\lambda }\\ $ Clearly, wavelength of the light used can be derived as: $\\\lambda =\dfrac{{{a}^{2}}}{Z} \\\Rightarrow \lambda =\dfrac{{{({{10}^{-2}})}^{2}}}{2.2} \\\Rightarrow \lambda =0.45\times {{10}^{-4}}\text{ m} \\ $ Clearly, frequency of light used can be derived as: $\\\lambda =\dfrac{c}{\nu } \\\Rightarrow\nu =\dfrac{c}{\nu } \\\Rightarrow\nu =\dfrac{3\times {{10}^{8}}}{0.45\times {{10}^{-4}}} \\\therefore\nu =6.66\times {{10}^{12}}\text{ Hz}$ 

 If the ratio of sine of angular spread of fourth secondary minimum and that of third secondary maximum of two respective diffraction responses of different wavelengths through the same opening is $2:3$ and if the sum of these two wavelengths is $10000\overset{\text{o}}{\mathop{\text{A}}}\,$, what are these wavelengths? 

 Utilize the expressions for ${{n}^{th}}$ secondary minimum as well as ${{n}^{th}}$ secondary maximum with respect to diffraction through a single slit to obtain a ratio of wavelengths used. Further use algebra to arrive at the exact values of these wavelengths. 

 In diffraction through a single slit, for ${{n}^{th}}$ secondary minimum, $\\a\sin {{\theta }_{n}}=n\lambda\\$ Also, for ${{n}^{th}}$ secondary maximum, $\\a\sin {{\theta }_{n}}=(2n+1)\dfrac{\lambda }{2}\\ $ Ratio of sine of angular spread of fourth secondary minimum and that of third secondary maximum of two respective diffraction responses of different wavelengths can be expressed as: $\\\dfrac{a\sin {{\theta }_{4}}}{a\sin {{\theta }_{3}}}=\dfrac{4{{\lambda }_{1}}}{\left( (3\times 2+1)\dfrac{\lambda {}_{2}}{2} \right)} \\\Rightarrow \dfrac{\sin {{\theta }_{4}}}{\sin {{\theta }_{3}}}=\dfrac{8}{7}\left( \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}} \right) \\\Rightarrow \dfrac{2}{3}=\dfrac{8}{7}\left( \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}} \right) \\\Rightarrow \dfrac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\dfrac{14}{24}\text{}\to (1) \\ $ It is also given that $ \\{{\lambda }_{1}}+{{\lambda }_{2}}=10000\times {{10}^{-10}}\text{ m} \\\Rightarrow{{\lambda }_{1}}=(10000\times {{10}^{-10}}-{{\lambda }_{2}})\text{ m}\to \text{(2)} \\ $ Substituting $(2)$ in $(1)$, $\\\dfrac{14}{24}=\dfrac{10000\times {{10}^{-10}}}{{{\lambda }_{2}}}-1 \\\Rightarrow \dfrac{10000\times {{10}^{-10}}}{{{\lambda }_{2}}}=\dfrac{38}{24} \\\Rightarrow {{\lambda }_{2}}=\dfrac{240000\times {{10}^{-10}}}{38} \\\Rightarrow {{\lambda }_{2}}=6315.7\times {{10}^{-10}}\text{ m} \\\Rightarrow {{\lambda }_{1}}=(10000-6315.7)\times {{10}^{-10}}\text{ m} \\\therefore {{\lambda }_{1}}=3684.3\times {{10}^{-10}}\text{ m}$ 