Question

# A 2 cm tall object is placed perpendicular to the principal axis of a convex lens with a focal length of 10 cm. The distance of the object from the lens is 15 cm. Find the position, nature, and size of the image. Calculate the magnification of the lens.

• 1.2
• 2.-2
• 3.4
• 4.-4
Option 2:-2
Solution

A 2 cm tall object is placed perpendicular to the principal axis of a convex lens with a focal length of 10 cm. The distance of the object from the lens is 15 cm. The position and nature of the image are real and inverted and the size of the image is -4 cm. The magnification of the lens is negative as the image is real and inverted.

### Steps to Calculate Magnification of Lens

Step 1 - It is given that:

Focal length f = + 10 cm

Distance of object u = - 15 cm

Height of object h = + 2 cm

Now we have to find Image distance, v and Image height, h′

Step 2 - To find the image position:

From the lens formula, 1/v - 1/u = 1/f

Substituting the values, we get:

1/v = 1/10 - 1/15

On simplifying:

v = 30 cm

The image is created on the right side of the lens at 30 cm, as indicated by the positive sign of v. As a result, the image is accurate yet upside-down.

Step 3 - Determine the magnification:

Magnification:

m = h/h = v/u

Substituting the values, we get:

h/2.0 = +30/-15 = -2

On simplifying:

h = -2 x 2 = -4 cm

Magnification, m = v/u = 30/-15 = -2

The image is inverted and actual, as indicated by the negative sign with the image's height and magnification. Thus, on the right side of the lens at 30 cm, a genuine image with a height of 4 cm is created. Double the size of the actual thing, and the image is inverted.

Summary:

A 2 cm tall object is placed perpendicular to the principal axis of a convex lens with a focal length of 10 cm. The distance of the object from the lens is 15 cm. Find the position, nature, and size of the image. Calculate the magnification of the lens.

An object that is 2 cm tall is positioned perpendicular to the primary axis of a convex lens that has a 10 cm focal length. 15 centimeters separate the item from the lens. The image is actual and reversed, with a real-world position, and it is -4 cm in size. Since the image is actual and reversed, the lens's magnification is negative. write a comment
Prev. Question
Next Question
Similar Questions
View All
SimplificationArithmetic Get solutions to similar questions at one place    GradeStack Learning Pvt. Ltd.Windsor IT Park, Tower - A, 2nd Floor, Sector 125, Noida, Uttar Pradesh 201303 help@byjusexamprep.com