A point object is placed at a distance of 12 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: Object distance $u = -12 \, cm$,Image distance $v = 18 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substitutin

Vedclass · Accepted Answer

Focal length = $7.2 \, cm$,Yes,magnified

Vedclass · Answer

(A) Given: Object distance $u = -12 \, cm$,Image distance $v = 18 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{f} = \frac{1}{18} - \frac{1}{-12} = \frac{1}{18} + \frac{1}{12}$.Taking the $LCM$ of $18$ and $12$ as $36$: $\frac{1}{f} = \frac{2+3}{36} = \frac{5}{36}$.Therefore,$f = \frac{36}{5} = 7.2 \, cm$.The magnification $m$ is given by $m = \frac{v}{u} = \frac{18}{-12} = -1.5$.Since the magnitude of magnification $|m| = 1.5 > 1$,the image is magnified.

$A$ point object is placed at a distance of $12 \, cm$ from a convex lens on its principal axis. Its image is formed on the other side of the lens at a distance of $18 \, cm$ from the lens. Find the focal length of the lens. Is the image magnified? Justify your answer.

Similar Questions

If you want to see an enlarged image of your face, which type of mirror will you use? Where will you place your face?

Give the mathematical expression for the power of a lens.

Draw ray diagrams showing the image formation by a convex lens when an object is placed:
$(a)$ Between optical centre and focus of the lens.
$(b)$ Between focus and twice the focal length of the lens.
$(c)$ At twice the focal length of the lens.
$(d)$ At infinity.
$(e)$ At the focus of the lens.

The radius of curvature of a spherical mirror is half of the focal length.

$A$ convex lens of focal length $10 \ cm$ is placed at a distance of $12 \ cm$ from a wall. How far from the lens should an object be placed so as to form its real image on the wall (in $cm$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module