A 2.0 text{ cm} tall object is placed perpend | vedclass

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

Question

(A) Given: Object height $O = 2.0 	ext{ cm}$,Focal length $f = +10 	ext{ cm}$,Object distance $u = -15 	ext{ cm}$.Using the lens formula $\frac{1}{

Vedclass · Accepted Answer

v = $30$ cm,Real,Inverted,$4$ cm

Vedclass · Answer

(A) Given: Object height $O = 2.0 	ext{ cm}$,Focal length $f = +10 	ext{ cm}$,Object distance $u = -15 	ext{ cm}$.Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{10} + \frac{1}{-15} = \frac{3 - 2}{30} = \frac{1}{30}$.Therefore,$v = +30 	ext{ cm}$. The positive sign indicates the image is formed on the other side of the lens.Magnification $m = \frac{v}{u} = \frac{I}{O}$.$I = \frac{v}{u} 	imes O = \frac{30}{-15} 	imes 2.0 = -4 	ext{ cm}$.The image is real,inverted,and $4 	ext{ cm}$ tall,formed at a distance of $30 	ext{ cm}$ from the lens.

$A$ $2.0 \text{ cm}$ tall object is placed perpendicular to the principal axis of a convex lens of focal length $10 \text{ cm}$. The distance of the object from the lens is $15 \text{ cm}$. Find the position,nature,and size of the image formed.

Similar Questions

Which of the following has the minimum refractive index?

An object $4 \ cm$ high is placed at a distance of $27 \ cm$ in front of a convex lens of focal length $18 \ cm$. Find the position,nature,and size of the image formed.

An object of height $5 \, cm$ is placed perpendicular to the principal axis of a concave lens of focal length $10 \, cm$. If the distance of the object from the optical centre of the lens is $20 \, cm$,determine the position,nature,and size of the image formed using the lens formula.

Vedclass Test Series

Exam Paper Generator

Online Exam Module