Question

an object 5 cm in length is held 25cm away a converging lens of focal length 10cm find the position size and nature of the image formed​

Answer 1

Answer:

position size is15

and image formed is 20

Explanation:

thanks

Answer 2

$\large\sf\underline\red{GIVEN:-}$

$\small\sf{u = -25cm}$

$\small\sf{h =5cm}$

$\small\sf{f = 10cm}$

$\large\sf\underline\green{A/C,}$

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

$\longrightarrow$$\small\sf{\frac{1}{v}=\frac{1}{f}+\frac{1}{u}}$

$\longrightarrow$$\small\sf{\frac{1}{10}-\frac{1}{25}}$

$\longrightarrow$$\small\sf{\frac{15}{250}}$

$\longrightarrow$$\small{\boxed{\sf{v=16.66cm}}}$

The positive value of v shows the image is formed at the other side of the lens.

$\longrightarrow$$\small\sf{m = - \frac{v}{u}}$

$\longrightarrow$$\small\sf{- \frac{16.66}{25}}$

$\longrightarrow$$\small{\boxed{\sf{-0.66}}}$

The negative sign shows that the image is real and formed behind the lens.

$\longrightarrow$$\small\sf{m = \frac{h'}{h}= \frac{h'}{5}}$

$\longrightarrow$$\small\sf{h' = m× h2}$

$\longrightarrow$$\small\sf{-0.66 × 5}$

$\longrightarrow$$\small{\boxed{\sf{-3.3cm}}}$

The negative value of image height indicates that the image formed is inverted.