Question

an object of 5cm height is placed at a distance of 15cm from a concave mirror of radius of curvature 20cm find the size of the image​

Answer 1

Answer:

We are given R = 20 cm. Then f = R/2 = 10 cm.

Use the mirror formula: 1/10 = 1/15 + 1/Di → Di = 30 cm.

The image distance being positive means that it is REAL. Real images are INVERTED.

The magnification is (30 / 15 = 2 x). The image is MAGNIFIED, and located on the same side as the object.

Since the mag is 2 X, the height of the image (its size) is: 5 x 2 = 10 cm.

Answer 2

Answer:

Given:

\sf{Radius \:of \:Curvature=-20cm}[/tex]

$\sf{\therefore Focal \:length=-10cm}$

$\sf{Object \:Distance=-15cm}$

$\sf{Mirror=Concave \:mirror}$

To Find:

$\sf{Distance \:of \:image \:from \:mirror}$

Solution:

$\sf{{\dfrac{1}{f}}={\dfrac{1}{v}}+{\dfrac{1}{u}}}$

$\sf{-{\dfrac{1}{10}}={\dfrac{1}{v}}-{\dfrac{1}{15}}}$

$\sf{{\dfrac{1}{v}}=-{\dfrac{1}{10}}+{\dfrac{1}{15}}}$

$\sf{{\dfrac{1}{v}}={\dfrac{-3+2}{30}}}$

$\sf{{\dfrac{1}{v}}=-{\dfrac{1}{30}}}$

$\sf{{v=-30cm}$

Thus, the image will be formed at 30cm from the pole of concave mirror and will be at the same side where the object is kept.