CBSE 2023 · Region 4 · Set 1 · Q35 · 4 marks
The lens maker's formula is useful to design lenses of desired focal lengths using surfaces of suitable radii of curvature. The focal length also depends on the refractive index of the material of the lens and the surrounding medium. The refractive index depends on the wavelength of the light used. The power of a lens is related to its focal length. Answer the following questions based on the above :(a)How will the power of a lens be affected with an increase of wavelength of light?(b)The radius of curvature of two surfaces of a convex lens is R each. For what value of $\displaystyle \mu$ of its material will its focal length become equal to R ?The focal length of a concave lens of $\displaystyle \mu=1 \cdot 5$ is $\displaystyle 20$ cm in air. It is completely immersed in water of $\displaystyle \mu=\frac{4}{3}$. Calculate its focal length in water.An object is placed in front of a lens which forms its erect image of magnification 3. The power of the lens is $\displaystyle 5$ D . Calculate the distance of the object and the image from the lens.
The lens maker's formula is useful to design lenses of desired focal lengths using surfaces of suitable radii of curvature. The focal length also depends on the refractive index of the material of the lens and the surrounding medium. The refractive index depends on the wavelength of the light used. The power of a lens is related to its focal length. Answer the following questions based on the above :
(a)
How will the power of a lens be affected with an increase of wavelength of light?
(b)
The radius of curvature of two surfaces of a convex lens is R each. For what value of $\displaystyle \mu$ of its material will its focal length become equal to R ?
The focal length of a concave lens of $\displaystyle \mu=1 \cdot 5$ is $\displaystyle 20$ cm in air. It is completely immersed in water of $\displaystyle \mu=\frac{4}{3}$. Calculate its focal length in water.
An object is placed in front of a lens which forms its erect image of magnification 3. The power of the lens is $\displaystyle 5$ D . Calculate the distance of the object and the image from the lens.
Marking-scheme solution
(a) \(\displaystyle P=\frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\[\mu \propto \frac{1}{\lambda} \Rightarrow P \propto \frac{1}{\lambda}
\]
So P decreases as \(\displaystyle \lambda\) increases.(b) \(\displaystyle \frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\[\frac{1}{R}=(\mu-1) \frac{2}{R} \quad\left(R_{1}=R, R_{2}=-R\right)
\]
(c) \(\displaystyle \frac{1}{f}=(\mu-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
\[\frac{1}{20}=\left(\frac{3}{2}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)
\] When the lens is immersed in water
\[\frac{1}{f_{\text {unknown }}}=\left(\frac{9}{8}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)
\] \(\displaystyle \frac{\dfrac{1}{20}}{\dfrac{1}{f}}=\frac{1 / 2}{1 / 8}=\Rightarrow f=80 \mathrm{~cm}\)
OR OR
\[m=3=\frac{v}{u} \Rightarrow v=3 u
\] \(\displaystyle f=\frac{100}{+5}=+20 \mathrm{~cm} \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
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