Question

a convex lens of focal length 40cm is placed in system of coordinate axis such that its optical centre is at origin and principal axis coinciding with x-axis. the coordinates of a point object in front of the lens is (-50cm, 5cm). The coordinates of image formed by lens is

A. (50cm, -5cm)
B. (-200cm, 10cm)
C. (200cm, -20cm)
D. (200cm, 20cm)

Answer 1

Answer:

This is the answer of your question

Answer 2

SOLUTION

TO CHOOSE THE CORRECT OPTION

A convex lens of focal length 40cm is placed in system of coordinate axis such that its optical centre is at origin and principal axis coinciding with x-axis. the coordinates of a point object in front of the lens is (-50cm, 5cm). The coordinates of image formed by lens is

A. (50cm, -5cm)

B. (-200cm, 10cm)

C. (200cm, -20cm)

D. (200cm, 20cm)

EVALUATION

Here by the given condition

f = Focal length = 40 cm

u = - 50 cm

We know that

$\displaystyle \sf{ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} }$

$\displaystyle \sf{ \implies \frac{1}{v} = \frac{1}{u} + \frac{1}{f} }$

$\displaystyle \sf{ \implies \frac{1}{v} = \frac{1}{ - 50} + \frac{1}{40} }$

$\displaystyle \sf{ \implies \frac{1}{v} = - \frac{1}{ 50} + \frac{1}{40} }$

$\displaystyle \sf{ \implies \frac{1}{v} = \frac{ - 4 + 5}{ 200} }$

$\displaystyle \sf{ \implies \frac{1}{v} = \frac{ 1}{ 200} }$

$\displaystyle \sf{ \implies v = 200 }$

$\boxed{ \: \: \displaystyle \sf{ v = 200 \: \: cm} \: \: }$

Now

h = Height of the object = 5 cm

h' = Height of the image = ?

We know that

$\displaystyle \sf{Linear \: magnification = m = \frac{Height \: of \: the \: image}{Height \: of \: the \: object} = \frac{h'}{h} }$

Also linear magnification is related to Image distance and Object distance as below

$\displaystyle \sf{Linear \: magnification = m = \frac{Image \: distance}{Object \: distance} = \frac{v}{u} }$

Thus we get

$\displaystyle \sf{ m = \frac{h'}{h} = \frac{v}{u} }$

$\displaystyle \sf{ \implies \: \frac{h'}{5} = \frac{200}{ - 50} }$

$\displaystyle \sf{ \implies \: \frac{h'}{5} =- 4}$

$\displaystyle \sf{ \implies \: h' = - 4 \times 5}$

$\displaystyle \sf{ \implies \: h' =-20}$

Hence the coordinates of image formed by lens is ( 200 cm , - 20 cm )

FINAL ANSWER

Hence the correct option is

C. ( 200 cm , - 20 cm )

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