How do You Identify a Mirror as Concave or Convex Without Touching it?

Advanced Physics Topics like quantum mechanics and relativity have revolutionized our understanding of the universe.

What is a Spherical Aberration? What are the Uses of Spherical Mirrors?

Method Of Identifying Mirrors

If an object is placed in front of a plane mirror, a virtual, erect image of the same size as the object is formed. If an object is placed very close to a concave mirror a virtual, erect and magnified image is formed. A convex mirror forms a virtual, erect image smaller than the object.

For identification one can hold a pen or a finger very close to the mirror. If an erect image of the same size as the object is formed, then the mirror is a plane one. If an erect image larger than the object is formed, the mirror is concave. If it is smaller than the object the mirror is convex.

Nature of Image	Type of Mirror
Erect, same size as object	Plane
Erect, larger than the object	Concave
Erect, smaller than the object	convex

Spherical Aberration And Its Remedy

Spherical aberration: The mirror equation is applicable only for spherical mirrors of small aperture to increase the intensity of illumination of the reflected rays to get a brighter image, we often use spherical mirrors of large aperture. If a beam of rays is incident parallel to the principal axis of a concave mirror of large aperture not all the rays meet after reflection at a single point. Rather, the reflected rays F meet at various points of the principal axis between F to F₁ [Fig.].

So the spherical mirror image becomes indistinct. Larger the having more than one focus aperture of the mirror, the more indistinct is the image. This defect of the image is called spherical aberration.

Remedy for spherical aberration: If the shape of the mir-ror is changed from spherical to paraboloidal, it is possible to get an image free from spherical aberration.

Because, according to the geometrical having a single focus properties of the parabola, all the rays parallel to the principal axis of a paraboloidal mirror meet at its focus after reflection from it [Fig.].

Uses of Spherical Mirrors

Concave mirror:

1. Concave mirrors are often used as shaving glasses (mirrors) to see the magnified image of the face, its distance being less than the focal length of the mirrors.
2. Small concave mirrors are used by doctors to focus a parallel beam of light on the affected parts like eye, ear, throat etc. to examine them.
3. A small electric lamp placed at the focus of a concave mirror produces a parallel beam of light. So concave mirrors are used as reflectors in torches and car headlights. For better result paraboloidal mirror can be used as car headlights.
4. Concave mirrors are used in solar cookers.
5. The Palomar observatory in California has the best reflecting telescope which uses concave mirror for studying distant stars.

Convex mirror: Convex mirrors are used as rear-view mirrors in automobiles and other vehicles designed to allow the driver to see through the rear windshield. This is because a convex mirror forms erect and diminished images of objects and give a wider field of view compared to that for a plane mirror [Fig].

Paraboloidal mirror: According to the geometrical properties of the parabola, the rays parallel to the principal axis of a paraboloidal mirror meet at its focus after reflection from the mirror. So, if a source of light is placed at the focus F of a paraboloidal mirror, the reflected rays proceed ahead parallel to the principle axis [Fig.]. For this reason paraboloidal mirrors are used in car headlights and search-lights.

Sign rules that have followed here: To solve the numerical problems, a few sign conventions have followed here.

1. The value of u, v, f or r is used with their proper sign in the mirror equation, viz, \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\) = \(\frac{2}{r}\).
2. Focal length for concave mirror is considered as negative and that for convex mirror is positive.
3. If image distance is negative, it indicates that image is formed at the same side of the object, and the image is real and inverted. If image distance is positive, it indicates that image is formed behind the mirror, and the image is virtual and erect.
4. Though m = \(-\frac{v}{u}\) is the standard formula for any type of spherical mirror, to find u or u from this equation, we only take the mod value of m.
5. For determination of nature of the image, to find the value of m, the associated formula is used following proper sign convention of u and v. If m becomes positive, the image will be erect, and m becomes negative, the image will be inverted.

Numerical Examples

Example 1.
An object is placed 60 cm away from a convex mirror. The size of the image is \(\frac{1}{3}\)rd the size of the object. Determine the radius of curvature of the mirror. [HS(XI) ‘06]
Solution:
We have trom the mirror equation,
f = \(\frac{u v}{u+v}\) …… (1)
Since, the image of a real object formed by a convex mirror is virtual therefore, v is positive but u is negative [Fig.].

According to the question,
m = \(\left|-\frac{v}{u}\right|\) = \(+\frac{1}{3}\)
or v = \(\left|-\frac{u}{3}\right|\) = \(\left|\frac{60}{3}\right|\) [∵ u = 60 cm]
Now, substituting the values of u and v with their proper sign in equation (1) we get,
f = \(\frac{-60 \times(+20)}{-60+20}\) = +30 cm
∴ The radius of curvature of the spherical mirror,
r = 2f = 2 × (+30) = +60 cm

Example 2.
An object of size 5 cm is placed on the principal axis of a convex mirror at a distance of 10 cm from it. The focal length of the convex mirror is 20 cm. Determine the nature, position and size of the image formed. [HS ‘05]
Solution:
u = -10 cm; f = +20 cm
[∵ focal length of convex mirror is positive]
Mirror equation,
\(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
or, \(\frac{1}{v}\) = \(\frac{1}{f}\) – \(\frac{1}{u}\) = \(\frac{1}{20}\) + \(\frac{1}{10}\) = \(\frac{1+2}{20}\) = \(\frac{3}{20}\)
or, v = \(\frac{20}{3}\) = 6.67cm
From the positive sign, it can be inferred that the image is formed 6.67 cm behind the mirror. So the image is virtual.
Magnification, m = \(-\frac{v}{u}\) = \(-\frac{20 / 3}{-10}\) = \(\frac{2}{3}\)
As magnification is positive, so the image is erect.
∴ The sizeof the image = \(\frac{2}{3}\) × 5 = \(\frac{10}{3}\) = 3.33 cm.

Example 3.
The Image of an object placed 50 cm in front of a concave mirror is formed 2 m behind the mirror. Determine its principal focus and radius of curvature. [HS(XI) ‘04]
Solution:
Here, u = -50 cm; v = +2 m = +200 cm [since image is formed behind the mirror, v is positive]
We have, \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\)
or f = \(\frac{u v}{u+v}\) = \(\frac{-50 \times(+200)}{-50+(+200)}\) = \(\frac{-200}{3}\) = -66.67 cm
∴ r = 2f = 2 × \(\frac{-200}{3}\) = \(\frac{-400}{3}\) = -133.3 cm
So, the focal length of the concave mirror = 66.67cm and its radius of curvature = 135.3 cm

Example 4.
An object of length 5 cm is placed perpendicularly on the principal axis at a distance of 75 cm from a concave mirror. If the radius of curvature of the mirror is 60 cm, calculate the image distance and its height.
Solution:
Here, u = -75 cm; r = -60 cm;
∴ f = \(\frac{r}{2}\) = \(\frac{-60}{2}\) = -30 cm
We know, \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\) or \(\frac{1}{v}\) + \(\frac{1}{-75}\) = –\(\frac{1}{30}\)
or, \(\frac{1}{v}\) = –\(\frac{1}{30}\) + \(\frac{1}{75}\) = –\(\frac{1}{50}\) or, v = -50 cm
So, the image is formed at a distance of 50 cm in front of the mirror. The image is real.
Again, m \(=\frac{\text { height of the image }(I)}{\text { height of the object }(O)}\) = \(\frac{v}{u}\)
Here we take only the mod value of m as we no need to know the nature of image thus formed.
or, \(\frac{I}{5}\) = \(\frac{50}{75}\) = \(\frac{2}{3}\) or, I = 5 × \(\frac{2}{3}\) = 3.33 cm
So, height of the image = 3.33cm .

Example 5.
A beam of converging rays is incident on a convex mirror of focal length 30 cm. In the absence of the mirror the converging rays would meet at a distance of 20 cm from the pole of the mirror. If the mirror is situated at the said position where will the converging rays meet?
Solution:
In the absence of the mirror the converging rays would meet at P. So, P is the virtual object in case of the mirror [Fig.].

In case of virtual object, OP = u = +20 cm , f= +30 cm, image distance, v = ?
We know, \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{1}{v}\) + \(\frac{1}{20}\) = +\(\frac{1}{30}\)
or \(\frac{1}{v}\) = –\(\frac{1}{20}\) + \(\frac{1}{30}\) = \(\frac{-1}{60}\) or v = -60 cm
As v is negative, the converging rays will meet at Q at a distance of 60 cm in front of the mirror.

Example 7.
An image of size \(\frac{1}{n}\) times that of the object is formed in a convex mirror. If r is the radius of curvature of the mirror, calculate the object distance. [WBIEE 05]
Solution:
Considering the mod value only, magnification,
m = \(\frac{1}{n}\) = \(\frac{v}{u}\) or, v = \(\frac{u}{n}\)
We know, \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{1}{v}\) – \(\frac{4}{3 f}\) = –\(\frac{1}{f}\)
or, \(\frac{1}{v}\) = –\(\frac{1}{f}\) + \(\frac{4}{3 f}\) = +\(\frac{1}{3 f}\) or, v = +3f
The positive sign of v indicates that the image formed by the concave mirror is virtual in nature and is situated behind the mirror at a distance of 3f.
Magnification, m \(=\frac{\text { height of the image }}{\text { height of the object }}\)
or, 4 \(=\frac{\text { height of the image }}{2.5}\)
or height of the image = 10 cm

Example 8.
The image of the flame of a candle due to a mirror is formed on a screen at a distance of 9 cm from the can die. The image is magnified 4 times. Determine the nature, position and focal length of the mirror.
Solution:
As a magnified image is formed on a screen, the image is real.

Let u = -x cm and hence v = -(x + 9) cm
Here, m = 4 or, \(\frac{v}{u}\) = 4 [taking only the mod value of m]
or, \(\frac{x+9}{x}\) = 4 or, 4x = x + 9 or, x = 3
∴ u = -3 cm
So, the mirror is situated at a distance of 3 cm from the flame.
Now, v = -(3 + 9) = -12 cm
we know, \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{1}{-12}\) + \(\frac{1}{-3}\) = \(\frac{1}{f}\)
So, the focal length of the mirror is 2.4 cm. Its negative sign indicates the nature of the mirror as a concave one [Fig.].

Example 9.
The focal length of a concave mirror is f. A point object is placed beyond the focal length at a distance xf from the focus. Prove that the image will be formed beyond the focal length at a distance \(\frac{f}{x}\) from the focus and the magnification of the image will be (-\(\frac{1}{x}\)).
Solution:
Here, object distance, u = -(f + xf)

So, the image is formed beyond the focal length at a distance of \(\frac{f}{x}\) from the focus.
∴ Magnification, m = \(-\frac{v}{u}\) = \(-\frac{-\frac{f(1+x)}{x}}{-f(1+x)}\) = \(-\frac{1}{x}\)

Example 10.
A thin glass plate is placed in between a convex mirror of focal length 20 cm and a point source. The distance between the glass plate and the mirror is 5 cm. The image formed by the reflected rays from the front face of the glass plate and that due to the reflected rays by the convex mirror coincide at the same point. What is the distance of the glass plate from the source? Draw the ray diagram.
Solution:
In Fig. the ray diagram has been shown. The distance between the glass plate M and the point source P = x cm (say). Light rays starting from P form the image at Q by reflection at the convex mirror. By the question, the plane glass plate also forms the image of P at Q.

∴ Distance of the image from the glass plate = x cm
∴ Distance of the image from the convex mirror = (x – 5) cm
and distance of the object from the convex mirror = -(x + 5) cm
We know, \(\frac{1}{v}\) + \(\frac{1}{u}\) = \(\frac{1}{f}\) or, \(\frac{1}{x-5}+\frac{1}{-(x+5)}\) = +\(\frac{1}{20}\)
or, \(\frac{+x+5-x+5}{(x-5)(x+5)}\) = +\(\frac{1}{20}\) or \(\frac{10}{x^2-25}\) = +\(\frac{1}{20}\)
or, x² – 25 = 200 or, x² = 225 or, x = 15 cm
∴ The distance of the glass plate from the source = 15 cm.