
A concave lens of glass, refractive index $1.5$ has both surfaces of same radius of curvature R. On immersion in a medium of refractive index $1.75$, it will behave as a
a)convergent lens of focal length $3.5$ R
b)convergent lens of focal length $3.0$ R
c)divergent lens of focal length $3.5$ R
d)divergent lens of focal length $3.0$ R
Answer
533.4k+ views
Hint: Convergent and divergent nature of lens can be found by the sign of focal length. If F is positive then lens will be convergent and When F is negative then lens will be divergent. We will use the lens maker formula for finding F as it gives relation between focal length, radius of curvature and refractive index.
Formula Used:
We will use Lens Maker Formula given by-
$\dfrac{1}{F}=\left(\dfrac{\mu_{L}}{\mu_{S}}-1\right) \left(\dfrac{1}{R_{1}} - \dfrac{1}{R_{2}}\right)$
Where, $\mu_{L}$ is the refractive index of the lens.
$\mu_{S}$ is the refractive index of the surrounding medium.
$R_{1}$and $R_{2}$ are the radius of curvature.
F is the focal length.
Complete answer:
Given-
$R_{1}$= -R (its radius is in negative x-axis)
$R_{2}$= R (its radius is in positive x-axis)
$\mu_{L}$= $1.5$
$\mu_{S}$= $1.75$
Now we will use the lens maker formula.
$\dfrac{1}{F}=\left(\dfrac{\mu_{L}}{\mu_{S}}-1\right) \left(\dfrac{1}{R_{1}} - \dfrac{1}{R_{2}}\right)$
Now we will put values of $R_{1}$, $R_{2}$,$\mu_{L}$ and $\mu_{S}$.
After putting all these values we will get-
$\dfrac{1}{F}=\left(\dfrac{1.5}{1.75}-1\right) \left(\dfrac{1}{-R} - \dfrac{1}{R}\right)$
$\dfrac{1}{F}=\left(\dfrac{1.5}{1.75}-1\right) \left(\dfrac{-2}{R}\right) = \left(\dfrac{-0.25}{1.75}\right) \left(\dfrac{-2}{R}\right)$
$F = \left(\dfrac{-1.75}{0.25}\right)\left(\dfrac{-R}{2}\right) = 3.5 R$
F is positive with a value of $3.5$ R. So, the lens will behave as a convergent lens of focal length $3.5$ R.
Option(a) is correct.
Additional information:
The lens's focal length depends on the refractive index of the lens's material and radii of curvature of two coverings. Lens manufacturers use it to make particular power lenses from the glass of a given refractive index.
Note:
During Lens maker formula derivation, it is assumed that lens is thin. If there is some thickness of the lens, we cannot use the lens maker formula. There should be the same material on both sides of the lens. This is a limitation of the lens maker formula.
Formula Used:
We will use Lens Maker Formula given by-
$\dfrac{1}{F}=\left(\dfrac{\mu_{L}}{\mu_{S}}-1\right) \left(\dfrac{1}{R_{1}} - \dfrac{1}{R_{2}}\right)$
Where, $\mu_{L}$ is the refractive index of the lens.
$\mu_{S}$ is the refractive index of the surrounding medium.
$R_{1}$and $R_{2}$ are the radius of curvature.
F is the focal length.
Complete answer:
Given-
$R_{1}$= -R (its radius is in negative x-axis)
$R_{2}$= R (its radius is in positive x-axis)
$\mu_{L}$= $1.5$
$\mu_{S}$= $1.75$
Now we will use the lens maker formula.
$\dfrac{1}{F}=\left(\dfrac{\mu_{L}}{\mu_{S}}-1\right) \left(\dfrac{1}{R_{1}} - \dfrac{1}{R_{2}}\right)$
Now we will put values of $R_{1}$, $R_{2}$,$\mu_{L}$ and $\mu_{S}$.
After putting all these values we will get-
$\dfrac{1}{F}=\left(\dfrac{1.5}{1.75}-1\right) \left(\dfrac{1}{-R} - \dfrac{1}{R}\right)$
$\dfrac{1}{F}=\left(\dfrac{1.5}{1.75}-1\right) \left(\dfrac{-2}{R}\right) = \left(\dfrac{-0.25}{1.75}\right) \left(\dfrac{-2}{R}\right)$
$F = \left(\dfrac{-1.75}{0.25}\right)\left(\dfrac{-R}{2}\right) = 3.5 R$
F is positive with a value of $3.5$ R. So, the lens will behave as a convergent lens of focal length $3.5$ R.
Option(a) is correct.
Additional information:
The lens's focal length depends on the refractive index of the lens's material and radii of curvature of two coverings. Lens manufacturers use it to make particular power lenses from the glass of a given refractive index.
Note:
During Lens maker formula derivation, it is assumed that lens is thin. If there is some thickness of the lens, we cannot use the lens maker formula. There should be the same material on both sides of the lens. This is a limitation of the lens maker formula.
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