Question

An object is placed infront of a concave mirror of radius of curvature 20cm at a distance of 5cm. Find the position,nature and magnification.

Answer 1

Answer:

Mirror Formula: 1/v + 1/u = 1/f

According to the question,

• u = -5 cm
• f = -20 cm
• v = ?

Applying Mirror Formula we get,

⇒ 1/v - 1/5 = -1/20

⇒ 1/v = 1/5 - 1/20

⇒ 1/v = 4/20 - 1/20

⇒ 1/v = 3/20

⇒ v = +20/3

⇒ v = +6.67 cm

Therefore image formed is 6.67 cm from the mirror. Since it has a positive sign, the image is virtual and erect.

Magnification = -v/u

Magnification = -6.67/-5 = +1.33

Since Magnification is greater than 1 it is enlarged image.

Therefore Image is 6.67 cm from the mirror and nature is virtual, erect and enlarged.

Answer 2

$\huge{\mathfrak{\red{\underline{Answer :-}}}}$

Given :-

Distance of object from mirror(u) = -5 cm

Focal length(f) = -20 cm

To Find :-

Distance of image from mirror(v) = ?

Solution :-

Using Mirror Formula we get,

We know that,

$\huge{\red{\boxed{\boxed{\bf{\frac{1}{v} - \frac{1}{u} = \frac{1}{f}}}}}}$

_______________[Put Values]

⇒ 1/v - 1/5 = -1/20

⇒ 1/v = 1/5 - 1/20

Taking LCM

⇒ 1/v = 4/20 - 1/20

⇒ 1/v = 3/20

⇒ v = 20/3

⇒ v = 6.67 cm

When v is positive ,them the image is virtual and erect.

$\bf{\red{\boxed{\huge{Magnification = \frac{-v}{u}}}}}$

Magnification = -6.67/-5

Magnification = +1.33