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Question: A man holds a spherical shaving mirror of radius of curvature \(60\) cm, and focal length \(30\) cm,...

A man holds a spherical shaving mirror of radius of curvature 6060 cm, and focal length 3030 cm, at a distance of 1515 cm, from his nose. Find the position of the image, and calculate the magnification.

Explanation

Solution

Spherical shaving mirrors are concave mirrors. Concave mirrors have negative focal length. Use the mirror formula to find the position of the image and then use the formula of magnification.

Complete step by step answer:
Shaving mirrors are usually concave in nature because when the face is placed close to a concave mirror, then the concave mirror produces an erect and magnified image of the face.The radius of curvature is the radius of the sphere from which the mirror is cut. The distance from vertex to the centre of curvature is known as radius of curvature and radius of curvature is denoted by ‘R’.

The distance between the pole P of the concave mirror and the focus F is known as Focal length of the concave mirror. Radius of curvature is twice the focal length.
R=2f ⇒f=R2 ⇒R=−60cm ⇒f=−602 ⇒f=−30 cm R=2f \\\ \Rightarrow f=\dfrac{R}{2} \\\ \Rightarrow R=-60cm \\\ \Rightarrow f=\dfrac{-60}{2} \\\ \Rightarrow f=-30\,cm \\\

Radius of curvature: R=−60 cmR=-60\,cm, Focal length: f=−30 cmf=-30\,cm, let uu be the distance of object from the spherical mirror: u=−15 cmu=-15\,cm
Now using mirror formula: 1f=1u+1v\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}
1f=1−15+1v ⇒1−30=1−15+1v ⇒1−30+115=1v ⇒130=1v ∴v=30 cm \dfrac{1}{f}=\dfrac{1}{-15}+\dfrac{1}{v} \\\ \Rightarrow \dfrac{1}{-30}=\dfrac{1}{-15}+\dfrac{1}{v} \\\ \Rightarrow \dfrac{1}{-30}+\dfrac{1}{15}=\dfrac{1}{v} \\\ \Rightarrow \dfrac{1}{30}=\dfrac{1}{v} \\\ \therefore v=30\,cm \\\
So, the position of the image is 30 cm30\,cm from the spherical mirror.

Magnification is defined as the ratio of height of image to the height of object. It tells about the amount by which the object is magnified with respect to the object.
Formula of magnification:
m=−vu⇒m=−30−15 ∴m=2m=\dfrac{-v}{u} \Rightarrow m=\dfrac{-30}{-15} \\\ \therefore m=2
So, we can say that the image is formed 30 cm30\,cm behind the mirror and erect in nature and magnification is 2.

Note: The focal length of the concave mirror is always positive and focal length of convex mirror is always negative. The focal length of a plane mirror is infinite.A mirror is a reflective surface that bounces off light, producing either a real image or a virtual image. When an object is placed in front of a mirror, the image of the same object is seen in the mirror.