Question

an object is p;laced at a distance of 10 cmfrom a convex lens of focal length 12 cm. find the position and nature of image

Answer 1

Answer

• The image is real, inverted, enlarged and at -60 cm

Explanation

Given

• Object Distance = -10 cm
• Focal length = 12

To Find

• Position and nature of the image

Solution

● Here we are gonna use the lens formula as we are given the focal length, the image distance and the lens used

✭ Image Distance

→ 1/f = 1/v - 1/u

→ 1/f + 1/u = 1/v

→ 1/12 + 1/(-10) = 1/v

→ 5/60 - 6/60 = 1/v

→ 5-6/60 = 1/v

→ -1/60 = 1/v

→ Image Distance = -60

✭ Magnification

→ M = v/u

→ M = -60/-10

→ Magnificent = 6

Answer 2

Answer:

$\huge \bf \: Given$

• Object distance = -10 cm
• Focal length = 12 cm

$\huge \bf \: To \: find$

Position and nature of image

$\huge \bf \: Solution$

Here, we will use the lens formula for finding position.

$\huge \boxed { \bf \frac{1}{F} = \frac{1}{V} - \frac{1}{U} }$

$\huge \bf \: \frac{1}{f} + \frac{1}{u} = \frac{1}{v}$

$\sf \: \dfrac{1}{12} + \dfrac{1}{( - 10)} = \dfrac{1}{v}$

$\sf \: \dfrac{5}{60} - \dfrac{6}{60} = \dfrac{1}{v}$

$\sf \dfrac{ - 1}{60} = \dfrac{1}{v}$

$\huge \sf \: image \: distance = - 60$

Now,

Magnification

$\huge \bf \: M \: = \frac{v}{ u}$

$\sf \: m \: = \dfrac{ - 60}{ - 10}$

$\sf \: m = 6$

• Nature :- Real, Inverted and Enlarge
• Distance = -60 cm