The distance between an object and its image | vedclass

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

Question

(B) Given,magnification $m = 2$ and the image is real. For a real image,the distance between the object and the image is $D = |u| + |v| = 0.72 \ m$.Si

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given,magnification $m = 2$ and the image is real. For a real image,the distance between the object and the image is $D = |u| + |v| = 0.72 \ m$.Since $m = \frac{|v|}{|u|} = 2$,we have $|v| = 2|u|$.Substituting this into the distance equation: $|u| + 2|u| = 0.72 \ m \Rightarrow 3|u| = 0.72 \ m \Rightarrow |u| = 0.24 \ m$ and $|v| = 0.48 \ m$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$,where $v = 0.48 \ m$ and $u = -0.24 \ m$:$\frac{1}{f} = \frac{1}{0.48} - \frac{1}{-0.24} = \frac{1 + 2}{0.48} = \frac{3}{0.48} \Rightarrow f = 0.16 \ m$.When the object is moved $0.04 \ m$ towards the lens,the new object distance is $u' = -(0.24 - 0.04) = -0.20 \ m$.Using the lens formula again: $\frac{1}{v'} - \frac{1}{-0.20} = \frac{1}{0.16} \Rightarrow \frac{1}{v'} = \frac{1}{0.16} - \frac{1}{0.20} = \frac{5 - 4}{0.80} = \frac{1}{0.80} \Rightarrow v' = 0.80 \ m$.The new magnification is $m' = \frac{v'}{u'} = \frac{0.80}{-(-0.20)} = 4$.

The distance between an object and its image produced by a converging lens is $0.72 \ m$. The magnification is $2$. What will be the magnification when the object is moved by $0.04 \ m$ towards the lens?

Similar Questions

$A$ plane wave passes through a convex lens. The geometrical shape of the wavefront that emerges is

$A$ convex lens produces a real image $m$ times the size of the object. What will be the distance of the object from the lens?

The distance between the object and the screen is $100 \, cm$. $A$ lens produces an image on the screen when it is placed at either of two positions $40 \, cm$ apart. The power of the lens is (approximately): (in $, D$)

Let $u$ and $v$ be the distances of the object and the image from a lens of focal length $f$. The correct graphical representation of $u$ and $v$ for a convex lens when $|u|>f$,is

$A$ convex lens of focal length $25 \ cm$ and made of glass with refractive index $1.5$ is immersed in water. The absolute change in focal length of the glass is [Use refractive index of water = $\frac{4}{3}$] (in $cm$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module