Question

Where should an object be placed from a converging lens of focal length 20cm,
so as to obtain a real magnified image​

Answer 1

$\huge{\bf{\green{\mathfrak{Question:-}}}}$

$\bullet{\mapsto}$ Where should an object be placed from a converging lens of focal length 20cm, so as to obtain a real magnified image.

$\huge {\bf{\orange{\mathfrak{Answer:-}}}}$

$\bullet{\leadsto} \: \textsf{According to the question,}$

$\bullet{\leadsto} \: \textsf{\underline{Converging lens means Convex lens.}}$

$\bullet{\leadsto} \: \textsf{To obtain a real and magnified image,}$

$\bullet{\leadsto} \: \textsf{\underline{We have to place it between F and C.}}$

$\bullet{\leadsto} \: \textsf{F = 20 cm.}$

$\bullet{\leadsto} \: \textsf{C = ?}$

$\bullet{\leadsto} \: \sf C \: = \: 2f$

$\bullet{\leadsto} \: \sf C \: = \: 2 \: * \: 20$

$\bullet{\leadsto} \: \boxed{\sf C \: = \: 40 \: cm.}$

$\bullet{\leadsto} \: \textsf{In between F and C means,}$

$\bullet{\mapsto} \: \boxed{\therefore{\textsf{The object must be placed between 20 cm (F) and 40 cm (C).}}}$

$\huge{\bf{\purple{\mathfrak{Note :-}}}}$

$\bullet{\leadsto} \: \boxed{\textsf{See Attachment}}$

$\bullet{\leadsto} \: \textsf{Then,}$

$\bullet{\leadsto} \: \textsf{Characteristics of Image :-}$

• $\sf{Object \: placed \: :- \: Between \: F_{2} \: and \: C_{2}.}$

• $\textsf{It is a real image.}$

• $\textsf{It is inverted.}$

• $\sf{Image \: formed \: :-\: Beyond \: C_{1}.}$

• $\textsf{Object size < Image size i.e it is magnified.}$

$\huge{\bf{\blue{\mathfrak{Extra \: Information:-}}}}$

$\bullet{\leadsto} \: \textsf{Focal point (F) :-}$

• If a beam of light rays which are parallel to the principal axis of a convex lens then the rays will converge to a point. That point is called as Focus of a convex lens.

• If a beam of light rays which are parallel to the principal axis of a concave lens then the rays will diverge. If the diverged rays are extended backwards then they appear to come from a point. That point is called as the Focus of a concave lens.

$\bullet{\leadsto} \: \textsf{Converging lens means Convex lens.}$

$\bullet{\leadsto} \: \textsf{Diverging lens means Concave lens.}$

$\bullet{\leadsto} \: \textsf{Centre of Curvature (C) :-}$

• It is the centre of the sphere which the lens forms a part of it.

Answer 2

Given :-

Focal length = 20 cm

Solution :-

We know that m = v/u

magnified image = 2m

2 = v/u

v = 2u [1]

By using the mirror formula

1/f = 1/u + 1/v

1/f = 1/v - 1/u

1/20 = 1/2u - 1/u

1/20 = 1 - 2/2u

1/20 = -1/2u

u = -10 cm

v = 2(-10)  = -20 cm