Question

b). An object is placed in front of a convex lens such that the image formed has the same size as that of
the object. Draw a ray diagram to illustrate this.
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Answer 1

Answer:

when the object is on 2f the image is of same size

Answer 2

GiveN :-

• A object is placed in front of a convex lens .

To FinD :-

• At what distance from a convex lens should an object be placed on the axis so as to form a real image of the same size? Illustrate your with a ray diagram .

SolutioN :-

Here we need to find at what distance from a convex lens should an object be placed on the axis so as to form a real image of the same size. So let us assume that the focal length of the lens be f . And the height of object = height of the image = h .

For this the object should be placed at 2F , in order to obtain a image of same height . For the ray diagram refer to the attachment .

$\rule{200}{2}$

• P R O O F :-

$\red{\frak{Given}}\begin{cases}\textsf{ Height of the image =\textbf{h} . } \\\textsf{ Height of the object =\textbf{h} . } \\\textsf{ Focal length of the lens be =\textbf{f} . } \end{cases}$

According to the sign Convention we know that the focal length of a convex lens is positive . So ,

$\purple{\bigstar}\underline{\sf \mathscr{U} sing \ the \ \mathscr{M}agnification \ \textscr{F}ormula :- }$

$\sf:\implies\pink{ \dfrac{ Height_{(image)}}{Height_{(object)}}=-\dfrac{Distance_{(image)}}{Distance_{(object)}} }\\\\\sf:\implies \dfrac{-h}{h}=\dfrac{v}{-f} \\\\\sf:\implies -1 =-\dfrac{v}{-f} \\\\\sf:\implies\underset{\blue{\tt Hence \ Proved }}{\underbrace{ \boxed{\pink{\frak{Image \ Distance=- f }}}}}$