An object of height 5 , cm is placed perpendi | vedclass

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

Question

(A) Given: Height of object $h = +5 \, cm$,Focal length of concave lens $f = -10 \, cm$,Object distance $u = -20 \, cm$.Using the lens formula: $\frac

Vedclass · Accepted Answer

Position: $6.67 \, cm$ from lens,Nature: Virtual and erect,Size: $1.67 \, cm$

Vedclass · Answer

(A) Given: Height of object $h = +5 \, cm$,Focal length of concave lens $f = -10 \, cm$,Object distance $u = -20 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{v} = \frac{1}{f} + \frac{1}{u} = \frac{1}{-10} + \frac{1}{-20} = \frac{-2 - 1}{20} = \frac{-3}{20}$.Therefore,$v = -\frac{20}{3} \, cm \approx -6.67 \, cm$. The negative sign indicates the image is formed on the same side as the object.Nature: Since $v$ is negative,the image is virtual and erect.Size of image $(h')$: Using magnification $m = \frac{h'}{h} = \frac{v}{u}$.$h' = \frac{v}{u} 	imes h = \frac{-20/3}{-20} 	imes 5 = \frac{1}{3} 	imes 5 = +\frac{5}{3} \, cm \approx +1.67 \, cm$.

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.

Similar Questions

Copy the given figure and complete it to show the formation of the image of the object $AB$. Name the lens used in the figure.

Two lenses of power $+3.5 D$ and $-2.5 D$ are placed in contact. Find the power and focal length of the lens combination.

Is it possible for a lens to act as a convergent lens in one medium and a divergent lens in another?

Give two uses of a convex lens.

$(a)$ Two lenses have powers of $(i) +2 \text{ D}$ and $(ii) -4 \text{ D}$. What is the nature and focal length of each lens?
$(b)$ An object is kept at a distance of $100 \text{ cm}$ from a lens of power $-4 \text{ D}$. Calculate the image distance.

Vedclass Test Series

Exam Paper Generator

Online Exam Module