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

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(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.

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