Question

For two different positions of a thin convex lens between a real thin,erect object and screen,the length of images formed are 18cm and 8cm respectively. The lenght of the object is 1) 10cm 2) 12cm 3) 13cm 4) 14cm

Answer 1

Answer:

Explanation:

Here we have to use the concept of conjugate points. Conjugate points are those two positions of a lens between an object and screen wherein the object distance in one position becomes image distance for the other and vice versa.

For example, if in 1st position, u=15 cm and v=30cm

In 2nd position of the lens, if u=30cm, v will be equal to 15cm

And the magnification produced is

m=height of the image/height of the object.

And for conjugate points, magnification produced in 1st position×magnification produced in 2nd position=1

So here, let height of the object be h.

m1=18/h.....(1)

m2=8/h......(2)

m1×m2=1

So 144=h^2

h=12cm

Answer 2

The length of the object is 12 cm.

(2) is correct option.

Explanation:

Given that,

The length of images formed are 18 cm and 8 cm respectively.

Suppose, the total length is L.

Case (I),

$u = u_{1}$

$v=L-u_{1}$

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}=\dfrac{v_{i}}{u_{o}}$

Put the value into the formula

$\dfrac{L-u_{1}}{u_{1}}=\dfrac{8}{u_{0}}$.....(I)

Case (II),

$u=u_{2}$

$v=L-u_{2}$

We need to calculate the magnification

Using formula of magnification

$m=\dfrac{v}{u}=\dfrac{v_{o}}{u_{o}}$

Put the value into the formula

$\dfrac{L-u_{2}}{u_{2}}=\dfrac{18}{u_{0}}$.....(II)

The total length is

$L=u_{1}+u_{2}$

Put the value of L in equation (I)

$\dfrac{u_{1}+u_{2}-u_{1}}{u_{1}}=\dfrac{8}{u_{o}}$

$\dfrac{u_{2}}{u_{1}}=\dfrac{8}{u_{0}}$.....(III)

Now, put the value of L in equation (II)

$\dfrac{u_{1}+u_{2}-u_{2}}{u_{2}}=\dfrac{18}{u_{o}}$

$\dfrac{u_{1}}{u_{2}}=\dfrac{18}{u_{o}}$.....(Iv)

We need to calculate the length of object

From equation (III) and (IV)

$\dfrac{u_{2}}{u_{1}}\times\dfrac{u_{1}}{u_{2}}=\dfrac{8}{u_{0}}\times\dfrac{18}{u_{0}}$

$u_{0}^2=8\times18$

$u_{0}=\sqrt{144}$

$u_{0}=12\ cm$

Hence, The length of the object is 12 cm.

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Topic : optic

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