
A zone plate of focal length 60 cm, behaves as a convex lens, if wavelength of incident light is $6000A°$, then radius of first half period zone will be
$A. 36\times { 10 }^{ -8 }m$
$B. 6\times { 10 }^{ -8 }m$
$C. \sqrt {6} \times { 10 }^{ -8 }m$
$D. 6\times { 10 }^{ -4 }m$
Answer
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Hint: To solve this problem, use the Fresnel’s theory of diffraction. It gives the relationship between radius of half period zone and focal length of the zone plate. Substitute the values in the formula for order of period zone, wavelength of the incident light and focal length of the zone plate. Solve the equation and obtain the value of r which is the radius of half period zone.
Formula used:
${f}_{p}= \dfrac { { r }^{ 2 } }{ \left( 2p-1 \right) \lambda }$
Complete answer:
Given: Wavelength $\lambda= 6000A°= 6000 \times {10}^{-10}m$
Focal length of the zone plate ${f}_{p}$= 60 cm= 0.6 m
From Fresnel’s theory of diffraction, the relationship between radius of half period zone and focal length of the zone plate is given by,
${f}_{p}= \dfrac { { r }^{ 2 } }{ \left( 2p-1 \right) \lambda }$ …(1)
Where, p is the number of period zone
r is the radius of half period zone
$\lambda$ is the wavelength of the light
We have to find the first half period zone, thus, p= 1.
Substituting p=1 in the equation. (1) we get,
${ f }_{ 1 }=\dfrac { { r }^{ 2 } }{ \lambda }$
$\Rightarrow r= \sqrt { { f }_{ 1 }\lambda }$
Substituting values in above equation we get,
$r= \sqrt { 0.6\times 6000\times { 10 }^{ -10 } }$
$\Rightarrow r= \sqrt{36 \times {10}^{-8}}$
$\Rightarrow r= 6 \times {10}^{-4}$
Thus, the radius of the first half period zone will be $6 \times {10}^{-4}$.
So, the correct answer is option D i.e. $6\times { 10 }^{ -4 }m$.
Note:
Students must take care of the units first. They should convert the unit of every parameter to their S.I. units. Conversion of the units to the standard form should be carefully done as lack of accuracy and attentiveness here, may cause calculation error and thereby lead to wrong results. Students should remember that the radii of half period zones are proportional to the square root of the natural numbers.
Formula used:
${f}_{p}= \dfrac { { r }^{ 2 } }{ \left( 2p-1 \right) \lambda }$
Complete answer:
Given: Wavelength $\lambda= 6000A°= 6000 \times {10}^{-10}m$
Focal length of the zone plate ${f}_{p}$= 60 cm= 0.6 m
From Fresnel’s theory of diffraction, the relationship between radius of half period zone and focal length of the zone plate is given by,
${f}_{p}= \dfrac { { r }^{ 2 } }{ \left( 2p-1 \right) \lambda }$ …(1)
Where, p is the number of period zone
r is the radius of half period zone
$\lambda$ is the wavelength of the light
We have to find the first half period zone, thus, p= 1.
Substituting p=1 in the equation. (1) we get,
${ f }_{ 1 }=\dfrac { { r }^{ 2 } }{ \lambda }$
$\Rightarrow r= \sqrt { { f }_{ 1 }\lambda }$
Substituting values in above equation we get,
$r= \sqrt { 0.6\times 6000\times { 10 }^{ -10 } }$
$\Rightarrow r= \sqrt{36 \times {10}^{-8}}$
$\Rightarrow r= 6 \times {10}^{-4}$
Thus, the radius of the first half period zone will be $6 \times {10}^{-4}$.
So, the correct answer is option D i.e. $6\times { 10 }^{ -4 }m$.
Note:
Students must take care of the units first. They should convert the unit of every parameter to their S.I. units. Conversion of the units to the standard form should be carefully done as lack of accuracy and attentiveness here, may cause calculation error and thereby lead to wrong results. Students should remember that the radii of half period zones are proportional to the square root of the natural numbers.
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