Question

An object of size 5.0 cm is placed at 15 cm in front of a convex lens of focal length 20 cm. Find position, size and nature of the image​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

Given Information:

$\sf:\longmapsto Size\ Of\ Object (O) = 5.0\:cm$

$\sf:\longmapsto u = - 15 \: (negative)$

$\sf:\longmapsto f = 20 \: (positive)$

At first we will find out the distance of image (v).

By using lens formula, we get:

$\sf:\longmapsto \dfrac{1}{v} - \dfrac{1}{u}=\dfrac{1}{f}$

Substituting the value in the formula, we get:

$\sf:\longmapsto \dfrac{1}{v} - \dfrac{1}{ - 15}=\dfrac{1}{20}$

$\sf:\longmapsto \dfrac{1}{v} + \dfrac{1}{15}=\dfrac{1}{20}$

$\sf:\longmapsto \dfrac{15 + v}{15v} = \dfrac{1}{20}$

$\sf:\longmapsto 20(15 + v)= 15v$

$\sf:\longmapsto 300 + 20v= 15v$

$\sf:\longmapsto 20v- 15v = - 300$

$\sf:\longmapsto 5v = - 300$

$\sf:\longmapsto v = - 60$

Therefore, the distance of image is 60 cm.

Now, magnification is calculated using the formula given below:

$\sf: \longmapsto m = \dfrac{I}{O} = \dfrac{v}{u}$

$\sf: \longmapsto \dfrac{I}{O} = \dfrac{v}{u}$

$\sf: \longmapsto \dfrac{I}{O} = \dfrac{ - 60}{ - 15}$

As v > u, I > O. So, the length of image is greater than the length of object.

So, the image will be formed in front of the mirror. The image is magnified and it is virtual.

$\textsf{\large{\underline{Note}:}}$

→ u = Distance of object.

→ v = Distance of image.

→ f = Focal length.

→ I = Size of Image.

→ O = Size of Object.

For convex lens:

1. u is negative.
2. v is positive for real image and negative for virtual image.
3. f is always positive.

For concave lens:

1. u, v and f are negative.
2. Numerical value of u is always greater than that of v.

Answer 2

$\textsf{\large{\underline{Question}:}}$

An object of size 5.0 cm is placed at 15 cm in front of convex lens of focal length 20 cm.

$\textsf{\large{\underline{Find}:}}$

i. Position of the image.

ii. Size of the image.

iii. Nature of the image.

$\textsf{\large{\underline{Solution}:}}$

According to the question :

---> Size of the object (O) = 5.0 cm,

---> distance of the object (u) = -15 (negative),

---> focal length of the image (f) = 20 (positive).

By using lens formula, we will solve the problem :

$\implies \: \frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Putting respective values in the formula,

$\implies \: \frac{1}{v} - \frac{1}{( - 15)} = \frac{1}{20}$

$\implies \: \frac{1}{v} + \frac{1}{15} = \frac{1}{20}$

$\implies \: \frac{1}{v} = \frac{1}{20} - \frac{1}{15}$

L.C.M of 20 and 15 = 60

$\implies \: \frac{1}{v} = \frac{(3 - 4)}{60}$

$\implies \: \frac{1}{v} = \frac{ - 1}{60}$

$\implies \: v = - 60$

Therefore, the distance of the image is 60 cm.

We know, formula of magnification is : -

$m = \frac{ \: I }{O} = \frac{v}{u}$

$\implies \: \frac{ \: I}{O} = \frac{ - 60}{ - 15} = 4$

So, the magnification is 4.

Since, v > u & I > O, the length of image will be greater than that of the object.

$\textsf{\large{\underline{Answer}:}}$

i. Position of image : Image is formed in front of the mirror.

ii. Size of the image : Image is magnified.

iii. Nature of the image : Image is virtual , i.e. erect.