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Question: A converging crown glass lens has a focal length 20 cm for violet rays. Its focal length for red ray...

A converging crown glass lens has a focal length 20 cm for violet rays. Its focal length for red rays is (μv=1.56,μr=1.53)(\mu_v = 1.56,\mu_r = 1.53)
(A) 20.82cm
(B) 21.13cm
(C) 22.85cm
(D) 24.85cm

Explanation

Solution

Focal length of a lens can be found using its radius of curvature R1{R_1} and R2{R_2} , and the refractive index of the material μ\mu . With this information lens maker formula can be used to find the focal length of the crown glass for the lens for red rays.

Formula used: We will be using the lens maker formula,
1f=(μ1)[1R11R2]\dfrac{1}{f} = (\mu - 1)\left[ {\dfrac{1}{{R_1}} - \dfrac{1}{{R_2}}} \right] , where ff is the focal length to be found, μ\mu is the refractive index of material used in the lens, R1{R_1} and R2{R_2} is the radius of curvature of both the surfaces of the lens.

Complete Step by Step solution
A crown glass lens is a lens made from crown glass, which has lower refractive index and dispersion when compared to the other lenses. We know that the focal length for any lens is found using the lens maker formula 1f=(μ1)[1R11R2]\dfrac{1}{f} = (\mu - 1)\left[ {\dfrac{1}{{R_1}} - \dfrac{1}{{R_2}}} \right] .
Substituting the values of fv=20cmf_v = 20cm , μv=1.56\mu_v = 1.56 in the lens maker equation for finding focal length of the given crown glass lens for violet colour.
1fv=(μv1)[1R11R2]\Rightarrow \dfrac{1}{{f_v}} = (\mu_v - 1)\left[ {\dfrac{1}{{R_1}} - \dfrac{1}{{R_2}}} \right]
120=(1.561)[1R11R2]\Rightarrow \dfrac{1}{{20}} = (1.56 - 1)\left[ {\dfrac{1}{{R_1}} - \dfrac{1}{{R_2}}} \right]
Now, similarly we can also rewrite the lens maker formula to find the focal length of the given crown glass lens for red colour.
1fr=(μr1)[1R11R2]\dfrac{1}{{f_r}} = (\mu_r - 1)\left[ {\dfrac{1}{{R_1}} - \dfrac{1}{{R_2}}} \right]
Substituting the value of μr=1.53\mu_r = 1.53 ,we get
1fr=(1.531)[1R11R2]\dfrac{1}{{f_r}} = (1.53 - 1)\left[ {\dfrac{1}{{R_1}} - \dfrac{1}{{R_2}}} \right]
You can note that the values of R1R_1 and R2R_2 remain the same because they are the radii of curvature of the same crown glass lens.
Let us divide the two equations, to find frfv\dfrac{{f_r}}{{f_v}} .
frfv=μv1μr1[1R11R21R11R2]\dfrac{{f_r}}{{f_v}} = \dfrac{{\mu_v - 1}}{{\mu_r - 1}}\left[ {\dfrac{{\dfrac{1}{{R_1}} - \dfrac{1}{{R_2}}}}{{\dfrac{1}{{R_1}} - \dfrac{1}{{R_2}}}}} \right]
The same terms in the numerator and denominator cancels out and we are left with, frfv=μv1μr1\dfrac{{f_r}}{{f_v}} = \dfrac{{\mu_v - 1}}{{\mu_r - 1}}
Now substitute the values, fv=20cmf_v = 20cm , μr=1.53\mu_r = 1.53 , and μv=1.56\mu_v = 1.56 .
fr20=1.5611.531\dfrac{{f_r}}{{20}} = \dfrac{{1.56 - 1}}{{1.53 - 1}}
Solving for frf_r we get
fr=20(0.560.53)\Rightarrow f_r = 20(\dfrac{{0.56}}{{0.53}})
fr=21.13207cm\Rightarrow f_r = 21.13207cm
So, the focal length of the given crown glass lens for red colour is fr=21.13f_r = 21.13 , which is option B.

Note
Now that we can see that the crown glass lens has different focal lengths for different colours, this is because the focal length depends on the refractive indices for lenses and the refractive index is in turn related with the inverse of the square of wavelength. (and wavelength varies for each colour)
μ1λ2\Rightarrow \mu \propto \dfrac{1}{{{\lambda ^2}}}
This does not hold true in case of mirrors, and hence the focal length of a mirror does not change with respect to its colour.