Question

An object is kept at a distance of 40cm from a diverging lens of focal length 20cm At what distance the image is
formed from the lens ? Show through the ray diagram..​

Answer 1

Answer:

★ Figure refer to attachment

Given

An object is placed at a distance of 40 cm from a concave lens of focal length 20cm • Find the distance of image from the lens.

To find

List the characteristics of image formed by lens nature, position, and size) • Draw ray diagram to justify your answer.

Solution

Object distance (u) = -40cm

Focal length (f) = - 20cm

Image distance (v) = ?

Appy lens formula

\begin{gathered}\implies\sf \dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f} \\ \\ \\ \implies\sf \dfrac{1}{v}-\dfrac{(-1)}{40}=\dfrac{-1}{20} \\ \\ \\ \implies\sf \dfrac{1}{v}+\dfrac{1}{40}=\dfrac{-1}{20} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{-1}{40}-\dfrac{1}{20} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{-1-2}{40} \\ \\ \\ \implies\sf \dfrac{1}{v}=\dfrac{-3}{40} \\ \\ \\ \implies\sf v=\cancel\dfrac{-40}{3} \\ \\ \\ \implies\sf v=-13.3cm\end{gathered}

⟹

v

1

−

u

1

=

f

1

⟹

v

1

−

40

(−1)

=

20

−1

⟹

v

1

+

40

1

=

20

−1

⟹

v

1

=

40

−1

−

20

1

⟹

v

1

=

40

−1−2

⟹

v

1

=

40

−3

⟹v=

3

−40

⟹v=−13.3cm

Hence, the image distance is -13.3cm

Nature of image

Nature = Virtual and erect

Size = Diminished

Position = b/w F1 and optical centre

Explanation:

Mark me as brainliest