Answer

Verified

21k+ views

**Hint:**We want to find the final position of image for that we must have the focal length of lens only then we can find the position of image. To find the focal length of a lens although we have two Plano-convex lenses, we have to find the focal length of a combination of these Lens. When we find combined focal length then simply we apply lens formula to find the distance of the image.

**Complete step by step solution:**

In the question the given data is

Radius of curvature of both lanes $14cm$ and refractive index of lens 1is ${n_1} = 1.5$ and lens 2 is ${n_2} = 1.2$

**Step 1:**We have to find the individual focal length of both lenses by using formula.

$\dfrac{1}{f} = \left( {n - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ ....................... (1)

Where $n \Rightarrow $ refractive index of lens

And ${R_1} \Rightarrow $ radius of curvature of first surface of lens

${R_2} \Rightarrow $ Radius of curvature of second surface of lens

$f \Rightarrow $ Focal length of lens

So we apply this formula for the first lens. For first lens

Refractive index ${n_1} = 1.5$

${R_1} = 14cm$

${R_2} = \infty $ Because second surface is plane in first lens

Apply formula from eq (1)

$ \Rightarrow \dfrac{1}{{{f_1}}} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{{14}} - \dfrac{1}{\infty }} \right)$

$ \Rightarrow \dfrac{1}{{{f_1}}} = \left( {1.5 - 1} \right)\left( {\dfrac{1}{{14}}} \right)$

Solving it

$ \Rightarrow \dfrac{1}{{{f_1}}} = \dfrac{1}{{28}}$

So the focal length of lens 1

$\therefore {f_1} = 28cm$

Similarly we find the focal length of the second lens. For lens 2

Refractive index is ${n_2} = 1.2$

${R_1} = \infty $

${R_2} = - 14cm$ Because ${R_2}$ is in left side of optical canter (by sign convention we take left side distance negative in sign and take right side distance positive )

Apply formula from eq(1)

$ \Rightarrow \dfrac{1}{{{f_2}}} = \left( {1.2 - 1} \right)\left( {\dfrac{1}{\infty } - \dfrac{1}{{ - 14}}} \right)$

$ \Rightarrow \dfrac{1}{{{f_2}}} = \left( {1.2 - 1} \right)\left( {\dfrac{1}{{14}}} \right)$

Solving it

$ \Rightarrow \dfrac{1}{{{f_2}}} = \dfrac{1}{{70}}$

So focal length of lens 2

${f_2} = 70cm$

**Step 2:**In this step we find the focal length of combination of both lenses

We have formula for find equivalent focal length of two lenses is given as

$\dfrac{1}{{{f_{eq}}}} = \dfrac{1}{{{f_1}}} + \dfrac{1}{{{f_2}}}$

Using this formula

$ \Rightarrow \dfrac{1}{{{f_{eq}}}} = \dfrac{1}{{28}} + \dfrac{1}{{70}}$

Solving it

$ \Rightarrow \dfrac{1}{{{f_{eq}}}} = \dfrac{1}{{20}}$

So the equivalent focal length of these lenses

$\therefore {f_{eq}} = 20cm$

Step 3

Now we can find the distance of image by using lens formula

$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

Where $u \Rightarrow $ Distance of object

And $v \Rightarrow $ Distance of image

$f \Rightarrow $ Focal length of lens

In question it is given $u = - 40cm$

$f = 20cm$

Simply apply above formula

$ \Rightarrow \dfrac{1}{{20}} = \dfrac{1}{v} - \dfrac{1}{{ - 40}}$

Rearranging it

$ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{20}} - \dfrac{1}{{40}}$

On solving

$ \Rightarrow \dfrac{1}{v} = \dfrac{1}{{40}}$

So the distance of image is

$\therefore v = 40cm$ Right side of lens.

**Option (B) is correct here.**

**Note:**Some time we get confuse why some time some distances take positive and some take negative so the answer is we use sign convention to understand sign convention see the diagram

We take all distances positive in the right side of the lens and take negative in the left direction.Similarly, in positive y direction take positive height and in negative y direction take negative.

Recently Updated Pages

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

The path difference between two waves for constructive class 11 physics JEE_MAIN

What is the difference between solvation and hydra class 11 chemistry JEE_Main

IfFxdfrac1x2intlimits4xleft 4t22Ft rightdt then F4-class-12-maths-JEE_Main

Three point particles of mass 1 kg 15 kg and 25 kg class 11 physics JEE_Main

Sodium chloride is purified by passing hydrogen chloride class 11 chemistry JEE_Main

Other Pages

When propyne is treated with aqueous H2SO4 in presence class 12 chemistry JEE_Main

If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

when an object Is placed at a distance of 60 cm from class 12 physics JEE_Main

Assertion Acidic character of group 16 hydrides increases class 11 chemistry JEE_Main

The reaction of Zinc with dilute and concentrated nitric class 12 chemistry JEE_Main

Which of the following brings about dry bleaching A class 11 chemistry JEE_Main