Reflection at Curved Mirrors
Original Scanned PDF – View Notes
Basic Optical Terms
Reversibility of Light
Laws of Reflection
i. The incident ray, reflected ray and the normal at the point of incidence all lie in the same plane.
ii. The angle of incidence is equal to the angle of reflection: i = r.
iii. A ray normally incident on a reflecting surface is reflected back along the same initial path.
Object Distance and Image Distance
The distance of an object from the mirror is called object distance and is denoted by u. The distance of the image from the mirror is called image distance and is denoted by v.
Real Object and Virtual Object
Page 3 illustrates the two ray arrangements below: a real object can produce a virtual image, while converging incident rays may be treated as a virtual object and can produce a real image.
Real Image and Virtual Image
| Real Image | Virtual Image |
|---|---|
| It is formed by the actual intersection of reflected or refracted rays. | It is formed by the virtual intersection of reflected or refracted rays. |
| It can be obtained on a screen. | It cannot be obtained on a screen. |
| It is inverted with respect to the object. | It is erect with respect to the object. |
Reflection at Curved Mirrors
Spherical Mirror
1. Concave Mirror
The source also calls it a converging mirror because parallel rays incident on it converge at a point after reflection.
2. Convex Mirror
The source calls it a diverging mirror because parallel incident rays diverge after reflection and appear to converge at a point behind the mirror. The source states that a virtual image is obtained by a convex mirror.
Terms Used in Spherical Mirrors
Relation Between Focal Length (f) and Radius of Curvature (R)
The source states that for a spherical mirror, both concave and convex, the focal length is half of the radius of curvature.
Proof for Concave Mirror
From the figure:
i = r [law of reflection]
r = α [alternate-angle relation shown in the source]
Hence the source uses the geometry to obtain CF = BF.
For a very small aperture, B lies very close to the pole P, so BF ≈ FP.
Therefore:
CF = FP
CP − FP = FP
R − f = f
R = 2f
The source states that the same relation is obtained similarly for a convex mirror.
Mirror Formula
where f is focal length, u is object distance and v is image distance.
1. Concave Mirror – Real Image
The source takes AP = u, A′P = v, FP = f and CP = 2f.
From similar triangles ΔAPB and ΔA′PB:
A′B′/AB = A′P/AP = v/u … (i)
From the second pair of similar triangles used in the source and the small-aperture approximation FN ≈ FP:
A′B′/AB = (v − f)/f … (ii)
Using (i) and (ii):
v/u = (v − f)/f
vf = uv − uf
uv = uf + vf
Dividing by uvf:
1/f = 1/u + 1/v
2. Concave Mirror – Virtual Image
The page takes AP = u, A′P = −v and FP = f.
Using the similar triangles shown in the source:
A′B′/AB = −v/u … (i)
and:
A′B′/AB = (−v + f)/f … (ii)
Equating the two and simplifying gives:
1/f = 1/u + 1/v
3. Convex Mirror – Virtual Image
The source takes AP = u, A′P = −v and FP = −f.
From the similar triangles shown:
A′B′/AB = −v/u … (i)
and:
A′B′/AB = (−f + v)/(−f) … (ii)
Combining the source equations again gives:
1/f = 1/u + 1/v
Linear Magnification
The source also defines it as the ratio of image distance to object distance and denotes it by m.
where I = image height (size), O = object height (size), v = image distance and u = object distance.
Why Is a Concave Mirror Used for Shaving?
The source answer states that the face should be kept nearer to the mirror than its focus. The resulting image is then magnified and erect, making the face easier to see while shaving.
Solved Numericals from the Scanned Notes
Numerical 1 – Erect Image of Magnification 3 by a Concave Mirror
Question: At what position should an object be placed in front of a concave mirror of radius of curvature 0.4 m so that an erect image of magnification 3 is produced?
R = 0.4 m
f = R/2 = 0.2 m
m = 3
The source uses −v/u = 3:
v = −3u
Mirror formula:
1/0.2 = 1/u + 1/(−3u)
5 = (1/u)(1 − 1/3)
5u = 2/3
u ≈ 0.133 m
Thus the object is placed approximately 0.13 m from the mirror.
Numerical 2 – Image Length of a 4 m Pole along a Convex-Mirror Axis
Question: A pole 4 m long is laid along the principal axis of a convex mirror of focal length 1 m. The end nearer the mirror is 2 m from it. Find the length of the image of the pole.
For the nearer end:
f = −1 m, u = 2 m
1/f = 1/u + 1/v
−1 = 1/2 + 1/v
1/v = −3/2
v = −2/3 m
For the farther end, u′ = 2 + 4 = 6 m:
−1 = 1/6 + 1/v′
v′ = −6/7 m
Image length = |v′ − v|
= |−6/7 + 2/3|
Image length = 4/21 m ≈ 0.190 m
Numerical 3 – Erect Image Three Times the Object, R = 36 cm
Question: An erect image three times the size of the object is obtained with a concave mirror of radius of curvature 36 cm. What is the position of the object?
R = 36 cm
f = 18 cm
I/O = 3 and the source writes v/u = −3, so v = −3u.
1/18 = 1/u + 1/(−3u)
1/18 = 2/(3u)
The handwritten source boxes u = 12 m.
The data and algebra on the page are written in centimetres, while the final handwritten unit is “m”. This unit inconsistency is explicitly preserved instead of silently changing the source.
Numerical 4 – Image Height in a Convex Mirror
Question: An object 10 cm high is placed in front of a convex mirror of focal length 20 cm and the object is 30 cm from the mirror. Find the height of the image.
O = 10 cm, f = −20 cm, u = 30 cm
1/f = 1/u + 1/v
−1/20 = 1/30 + 1/v
1/v = −5/60
The source then writes v = −12 m.
Using I/O = v/u in the handwritten working:
I/10 = −12/30
The boxed image height is written as −4 with the unit handwriting unclear, followed by the statement “0.04 m or 4…” on the page.
This problem contains inconsistent handwritten units on page 12. The source values have been reported transparently rather than silently normalized.
Numerical 5 – Focal Length when Image Equals Object Size
Question: Calculate the focal length of a concave mirror when an object placed at a distance of 40 cm makes an image equal to the size of the object.
u = 40 cm
m = 1
The source takes v = 40 cm.
1/f = 1/40 + 1/40 = 2/40
f = 20 cm
Numerical 6 – Image of a Metre Scale along the Axis of a Convex Mirror
Question: A metre scale is placed along the axis of a convex mirror of focal length 25 cm, its nearer end being 50 cm from the mirror. Calculate the size of the image formed.
f = −25 cm
Near end: u = 50 cm
Using the mirror formula, the source obtains the image position for the near end as 50/3 cm in magnitude.
Far end: u′ = 150 cm
The source obtains the second image position and subtracts the two image distances.
Image size = 100/21 cm ≈ 4.76 cm
Conceptual Question – Real Image by a Convex Mirror
Question: A convex mirror with radius of curvature 30 cm forms a real image 20 cm from the pole. Explain how this is possible and find whether the image is erect or inverted.
The source writes:
R = 30 cm
f = 15 cm
It explains the possibility by considering converging incident rays. In that case, the incident beam corresponds to a virtual-object arrangement and a real image can be formed by the convex mirror.
On page 14, the handwritten solution gives the converging-ray explanation and a diagram, but it does not clearly write a final separate “erect” or “inverted” answer in text. No unsupported classification has been added here.
Discussion
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