Question

An object of height 3.0 cm is placed perpendicular to the principal axis of a concave lens of focal length 10 cm. The image is formed at a distance of 4 cm from the lens. Which of the following(s) is/are correct for the given condition?

• A. Object distance from the mirror is <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{20}{3}$
• B. Height of the image is 1.8 cm
• C. Height of the image is - 1.8 cm
• D. Image obtained is virtual and erect

Answer 1

Answer: (A), (B), (D)

The correct options are
A Object distance from the mirror is <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> $\frac{20}{3}$
B Height of the image is 1.8 cm
D Image obtained is virtual and erect

Given: height of object $h=3cm$, image distance $v=-4cm$ and focal length $f=-10cm$ (concave lens)
Let $u$ be the object distance
Using lens formula $\frac{1}{f}$ = $\frac{1}{v}$ - $\frac{1}{u}$
$\frac{1}{-10}$ = $\frac{1}{-4}$ - $\frac{1}{u}$
$\frac{1}{u}$ = $\frac{1}{-4}$ + $\frac{1}{10}$
$\frac{1}{u}$ = $\frac{-5+2}{20}$
or u = $\frac{-20}{3}cm$
So, Object distance from the lens is $\frac{-20}{3}$ cm
Magnification m = $\frac{heightofimage}{heightofobject}$ = $\frac{v}{u}$
$\frac{heightofimage}{heightofobject}$ = $\frac{-4\times 3}{-20}$
$\frac{heightofimage}{3}$ = $\frac{3}{5}$
height of image $=1.8cm$.
Positive sign suggests that image is virual and erect.