Question

A 6 cm tall object is placed perpendicular to the principal axis of a convex
lens of focal length 25 cm. The distance of the object from the lens is 40
cm. Determine:
(a) Position of image
(b) Nature of image.
(c) Size of image formed.​

Answer 1

$\bold{\underline{\underline{Answer:}}}$

Position of image : At a distance of 66.67 cm on either side of the object.

Nature of image :It is a real image.

Size of image formed : - 10 cm

$\bold{\underline{\underline{Step\:-\:by\:-\:step\:explanation:}}}$

Convex Lens :-

Given :

• A 6 cm tall object is placed perpendicular to the principal axis, Object Height, $\bold{h_1}$ = 6 cm
• Focal length of the lens = 25 cm
• Distance of the object from the lens is 40 cm, Object distance, u = - 40 cm

To find :

• Position of image (v)
• Nature of image.
• Nature of image. Size of image formed $\bold{(h_2)}$

Solution :

From lens formula,

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

$\implies$ $\bold{\frac{1}{25}}$ = $\bold{\frac{1}{v}}$ - $\bold{\frac{1}{-40}}$

$\implies$ $\bold{\frac{1}{25}}$ = $\bold{\frac{1}{v}}$ + $\bold{\frac{1}{40}}$

$\implies$ $\bold{\frac{1}{25}}$ - $\bold{\frac{1}{40}}$ = $\bold{\frac{1}{v}}$

Either using LCM method or cross multiplying numerator of other fraction with denominator of another fraction and then multiply the denominator of both the fractions with each other.

[Note : Perform all these above operation just on the LHS]

$\implies$ $\bold{\frac{40-25}{1000}}$ = $\bold{\frac{1}{v}}$

$\implies$ $\bold{\frac{15}{1000}}$ = $\bold{\frac{1}{v}}$

$\implies$ $\bold{\frac{3}{200}}$ = $\bold{\frac{1}{v}}$

$\implies$ $\bold{\frac{200}{3}}$ = v

$\implies$ $\bold{66.67=v}$

•°• Image will be formed at 66.67 cm from the lens on the other side of the lens with respect to the object.

• Nature of image :

The nature of image is real.

• Magnification of the image :

We know that the magnification of an image is calculated using the formula :-

Magnification,

$\implies$$\bold{\frac{h_2}{h_1}}{={\bold{\frac{v}{u}}}}$

Block in the values,

$\implies$$\bold{\frac{h_2}{6}}{={\bold{\frac{66.67}{-40}}}}$

$\implies$ $\bold{h_2\:=\frac{66.67\times\:6}{-40}}$

$\implies$ $\bold{h_2\:=\frac{400.02}{-40}}$

$\implies$ $\bold{h_2\:=\:-\:10.0005}$

•°• Size of image formed = -10 cm.

Answer 2

ANSWER:-

Given:

A 6cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 25cm.

The distance of the object from the lens is 40cm.

To find:

•Position of image.

•Nature of image.

•Size of image formed.

Explanation:

We have,

•Height of the object,h1=6cm.

•Focal length of the convex mirror,f=25cm

•Object distance,u=-40cm.

Using the lens formula:

$= > \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\ \\ = > \frac{1}{25} = \frac{1}{v} - (\frac{1}{ - 40} )\\ \\ = > \frac{1}{25} = \frac{1}{v} + \frac{1}{40} \\ \\ = > \frac{1}{v} = \frac{1}{25} - \frac{1}{40} \\ \\ = > \frac{1}{v} = \frac{8 - 5}{200} \\ \\ = > \frac{1}{v} = \frac{3}{200} \\ (cross \: multiplication) \\ = > 3v = 200 \\ \\ = > v = \frac{200}{3} \\ \\ = > v = 66.66cm$

Now,

magnification of image:

$m = \frac{v}{u} \: \: \: \: \: or \: \: \: \: m = \frac{h1}{h2}$

Equating both we get;

$= > \frac{h2}{h1} = \frac{v}{u} \\ \\ = > \frac{h2}{6} = \frac{66.66}{ - 40} \\ (cross \: multiplication) \\ = > h2 \times - 40 = 66.66 \times 6 \\ \\ = > h2 = \frac{66.66 \times 6}{ - 40} \\ \\ = > h2 = \frac{66.66 \times 6 \times 100}{ - 40 \times 100} \\ \\ = > h2 = \frac{6666 \times 6}{ - 4000} \\ \\ = > h2 = - 1.665 \times 6 \\ \\ = > h2 = 9.99cm$

h2 = -10cm (approx) is size of image formed.

Nature of image:

Real image. (is always inverted).

Thank you.