Question

When object is placed in between convex lens and its focal point, image formed is
A. Diminishes, real and inverted
B. Same size, real and inverted
C. Same size, real and upright
D. Magnified, virtual and upright

Answer 1

Hint: Position of image formed can be found using lens formula. We can use a magnification formula to determine the size ratio of the image to object.
A ray diagram can also be used to determine the details about the image.
Formula Used:
Lens formula, ${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}$; Magnification, $m={\displaystyle \frac{v}{u}}={\displaystyle \frac{h^{\prime }}{h}}$

Complete answer:

Above diagram is for the Formation of an image by a convex lens when the object is placed between focal point and optical centre.
The object is placed between the optical centre of the convex lens and its focal point. Therefore, the magnitude of distance of object from the optical centre,
$0<\left|u\right|<\left|f\right|$
Where $u$ is the distance of the object and $f$ is the focal length of the convex lens.
The focal length of the convex lens is positive and object distance is taken negative as per sign convention. The lens formula is given by
${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}$
${\displaystyle \frac{1}{v}}={\displaystyle \frac{1}{f}}+{\displaystyle \frac{1}{u}}$
$\Rightarrow v={\displaystyle \frac{u.f}{f+u}}$
Where $v$ is the distance of the image.
Since ${\displaystyle \frac{f}{f+u}}>1$ because f is positive and u is negative so makes denominator less than numerator.
$\Rightarrow v>u$
Since $\left|u\right|=-u<f$
$\Rightarrow {\displaystyle \frac{-1}{u}}>{\displaystyle \frac{1}{f}}$
$\Rightarrow {\displaystyle \frac{1}{f}}+{\displaystyle \frac{1}{u}}>0$
From lens formula, we get
$\Rightarrow {\displaystyle \frac{1}{v}}>0$
$\Rightarrow v<0$
From this, we can conclude that the image is formed on the left side as image distance is negative.
Now magnification of image if h’ and h are heights of image and object respectively, $m={\displaystyle \frac{v}{u}}={\displaystyle \frac{h^{\prime }}{h}}$
Since $v>u\Rightarrow m>1$
$\Rightarrow$ Image is magnified.
Because $v,u<0$, magnification is positive.
$\Rightarrow$ Image is virtual and upright.

Hence, option D is correct.

Note:
In mirror as well as optical lens problems, the most important thing to consider is sign conventions. Therefore, students should understand the sign conventions properly and then proceed with the solution. Diagrams are very helpful in understanding the sign conventions.
Focal length of a convex lens is always taken positively.