Question

For a concave spherical mirror with radius of curvature of 4 cm, find the object distance if the image is formed at a distance of 6 cm.​

Answer 1

☆Concept :-

A concave mirror is a sprecial mirror which is bulged out and creates all types of images (Real & inverted, Virtual & erect). It is mirror which has a negative radius of curvature and a negative focal length.

Here, image distance can be +ve or -ve depending upon its position over the principal axis. If the image is formed on the same side of the mirror, it will be -ve and will be stated as a real and inverted image if the image is be formed behind the mirror, it will be +ve and will be stated as a virtual and erect image.

• Radius of Curvature - Line passing through the mirror and centre of the Curvature of the sphere of which the mirror is a part.

• Image distance - distance of the image from the reflecting surface of the mirror. It can be +ve or -ve depending upon its position

• object distance - distance of the object from the reflecting surface of the mirror. It is always -ve because the object is always kept on same side of the mirror.

☆ Understanding The Question :-

We are provided with the Radius of Curvature and image distance and we are asked to find out the object distance. For that well first find out the focal length which is half of the radius of curvature. Then we'll use Mirror formula to find out the required value.

☆ Formula :-

$\\ \leadsto \large\underline{ \boxed{ \pink{ \sf \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }}} \bigstar$

Where, v, u, and f denote image distance, object distance and focal length respectively.

Use proper sign conventions and substitute the values to get the required answer.

☆ Sign Conventions :-

It is given that the radius of curvature of this concave is 4 cm which means that the focal length will be 2cm because focal length is just the half of radius. For calculation, we know that radius and focal length of any Concave Mirror is -ve. Hence, We'll use the focal length as (-2cm)

Now, it is also given that the image formed at a distance of 6cm on same side of the lens I.e (-6cm). Which means that the image formed is real and inverted.

☆ Calculation :-

$\\ \quad \longrightarrow \sf \dfrac{1}{ - 6} + \dfrac{1}{u} = \dfrac{1}{ - 2} \\ \\ \quad \longrightarrow \sf \frac{1}{u} = - \frac{1}{2} + \frac{1}{6} \\ \\ \quad \longrightarrow \sf \frac{1}{u} = \frac{ - 3 + 1}{6} \\ \\ \quad \longrightarrow \sf \frac{1}{u} = \frac{ - 2}{6} = \frac{ - 1}{3} \\ \\ \quad \longrightarrow \boxed{ \boxed{ \orange{\sf u = - 3 cm }}} \bigstar$

$\\ \therefore \underline{ \sf object \: is \: kept \: at \: a \: distance \: of \: 3cm \: in \: front \: of \: mirror}$

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