Light - Reflection and Refraction-Exercise
Physics - Exercise
Q1. Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Solution:
Clay:
Explanation:
A lens is made from a transparent material that allows light to pass through and refract (bend) it
to form an image.
- Water, glass, and plastic are transparent substances, so they can be used to make lenses.
- Clay, however, is opaque — it does not allow light to pass through, so it cannot be used to make a lens.
Q2. The image formed by a concave mirror is observed to be virtual, erect and larger than the
object. Where should be the position of the object?
(a) Between the principal focus and the
centre of
curvature
(b) At the centre of curvature
(c) Beyond the centre of curvature
(d) Between
the pole of
the mirror and its principal focus.
Solution:
Option: d
Explanation:
The image formed by a concave mirror is virtual, erect, and larger than the object when the object
is placed between the pole of the mirror and its principal focus.
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| Object Position | Nature of Image | Orientation | Size | Correct Option |
|---|---|---|---|---|
| Between pole (P) and focus (F) | Virtual, behind mirror | Erect | Larger | (d) |
| Between focus and center of curvature | Real | Inverted | Larger | |
| At centre of curvature (C) | Real | Inverted | Same | |
| Beyond center of curvature | Real | Inverted | Smaller |
Q3. Where should an object be placed in front of a convex lens to get a real image of the size of the object?
(a) At the principal focus of the lens
(b) At twice the focal length
(c) At
infinity
(d)
Between the optical centre of the lens and its principal focus.
Solution:
(b) At twice the focal length
Explanation:
To obtain a real image that is exactly the same size as the object using a convex lens, the object
must be placed at a distance equal to twice the focal length from the lens, which is known as the
"2F" position.
Q4. A spherical mirror and a thin spherical lens have each a focal length of –15 cm. The mirror and
the lens are likely to be
(a) both concave.
(b) both convex
(c) the mirror is concave
and the lens
is convex.
(d) the mirror is convex, but the lens is concave.
Solution:
(a) both concave.
Explanation:
The mirror and the lens, each having a focal length of –15 cm, indicate:
- For spherical mirrors, a negative focal length means the mirror is concave.
- For thin spherical lenses, a negative focal length means the lens is concave (diverging).
5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.
Solution:
(d) either plane or convex
Explanatiom:
If a mirror always gives an erect image, no matter how far away the observer stands, that mirror must be either a plane mirror or a convex mirror.
6. Which of the following lenses would you prefer to use while reading small letters found in a
dictionary?
(a) A convex lens of focal length 50 cm.
(b) A concave lens of focal length 50
cm.
(c) A
convex lens of focal length 5 cm.
(d) A concave lens of focal length 5 cm.
Solution
(c) A convex lens of focal length 5 cm.
Explanation:
To read small letters in a dictionary, a lens that magnifies is needed. Magnification is best
achieved with a convex lens (converging lens), since it can create a virtual, enlarged image when
the object is placed closer to it than its focal length.
The degree of magnification increases as the focal length decreases. A shorter focal length convex
lens provides greater magnification, making small text most readable.
Q7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distances of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
Solution
To form an erect (upright), virtual image using a concave mirror, the object must be placed between
the pole (P) and the principal focus (F) of the mirror.
Object distance, \(0 \lt u \lt 15\ cm\)
Q8. Name the type of mirror used in the following situations.
(a) Headlights of a car.
(b)
Side/rear-view mirror of a vehicle.
(c) Solar furnace. Support your answer with a reason.
(a) Headlights of a Car
Type of Mirror: Concave Mirror
Reason:
Concave mirrors are used in car headlights because they can produce a strong, parallel beam of
light. When a bulb is placed at the focus of a concave mirror, the mirror reflects and concentrates
the diverging light rays into a parallel beam that can travel long distances and illuminate the road
ahead. This enhances visibility at night.
(b) Side/Rear-View Mirror of a Vehicle
Type of Mirror: Convex Mirror
Reason:
Convex mirrors are used for side/rear-view mirrors because they allow a wider field of view. The
convex shape causes parallel rays to diverge, forming an image that is diminished (smaller) and
upright, so a larger area can be seen behind the vehicle. This helps drivers spot cars and
pedestrians otherwise in blind spots.
(c) Solar Furnace
Type of Mirror: Concave Mirror
Reason:
A solar furnace utilises a large concave mirror to concentrate sunlight onto a small area (the
focus). The mirror collects parallel rays from the Sun and converges them at its focal point, where
the furnace is placed. This concentration of energy raises the temperature quickly for
high-efficiency heating.
| Application | Type of Mirror | Reason |
|---|---|---|
| Headlights of a car | Concave | Produces a parallel, concentrated beam for distant illumination |
| Side/rear-view mirror of a vehicle | Convex | Gives a wide field of view with a diminished, upright image |
| Solar furnace | Concave | Concentrates sunlight to a single spot for heating |
Q9. One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Solution
If one-half of a convex lens is covered with black paper, the lens still produces a complete image
of the object, but the image is fainter (less bright) than with the full lens uncovered.
Q10. An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Solution
An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm
Object Height \(h\) = 5 cm
Object Position \(u\) = -25 cm
Focal Length \(f\) of Lens = 10 cm
Using Lens Formula
\[\require{cancel}\begin{aligned}
\frac{1}{f}&=\frac{1}{v}-\frac{1}{u}\\\\\frac{1}{10}&=\frac{1}{v}-\left(\frac{1}{-25}\right)\\\\
\frac{1}{10}&=\frac{1}{v}+\frac{1}{25}\\\\\Rightarrow
\frac{1}{v}&=\frac{1}{10}-\frac{1}{25}\\\\
&=\frac{5-2}{50}\\\\&=\frac{3}{50}\\\\\Rightarrow v&=\frac{50}{3}\\\\v&\approx 16.67\ cm
\end{aligned}\]
Magnification of object \(m\)\[\begin{align}
m&=\frac{v}{u}\\\\&=\left(\frac{\left(\frac{50}{3}\right)}{-25}\right)\\\\&=-\frac{50}{3\times
25}\\\\&=-\frac{\cancelto{2}{50}}{3\times \cancelto{1}{25}}\\\\m&=-\frac{2}{3}\tag{i}
\end{align}\]
Magnification \(m\) can also be expressed as \[\
m=\frac{h'}{h}\]
where \(h'\) is height of image and \(h\) is height of Object, hence
\[\begin{align}
m&=\frac{h'}{h}\\\\&=\frac{h'}{5}\end{align}\] from equation (i) \[\begin{align}
m&=-\frac{2}{3}\\\\
-\frac{2}{3}&=\frac{h'}{5}\\\\\Rightarrow h'&=-\frac{10}{3}\\\\&\approx -3.33 cm
\end{align}\]
Image formed at a distance of 16.67 cm away from the lens.
Height of Image is 3.33 cm (-ve sign indicates that the image is inverted and smaller)
Q11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Solution
Given:
Focal length \(f\) of concave lens = −15 cm (negative for concave).
Image distance \(v\) = −10cm
(negative since images by concave lenses are always
virtual, thus on the same side as the object).
$$\begin{aligned}v&=-10cm\\
f&=-15cm\\
\dfrac{1}{f}&=\dfrac{1}{v}-\dfrac{1}{u}\\\\
\dfrac{1}{-15}&=\dfrac{1}{-10}-\dfrac{1}{u}\\\\
\Rightarrow \dfrac{1}{u}&=\dfrac{1}{-10}+\dfrac{1}{15}\\\\
&=\dfrac{1}{15}-\dfrac{1}{10}\\\\
&=\dfrac{2-3}{30}\\\\
&=-\dfrac{1}{30}\\\\
&u=-30\\\\
m&=\dfrac{v}{u}\\\\
&=\dfrac{-10}{30}\\\\
&=\dfrac{1}{3}\\\\
&=0.33cm\end{aligned}$$
Q12. An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Solution Position of object u =-10cm Focal length of the Mirror (f) = 15cm $$\begin{aligned}\dfrac{1}{f}&=\dfrac{1}{v}+\dfrac{1}{u}\\\\ \dfrac{1}{15}&=\dfrac{1}{v}+\left( \dfrac{1}{-10}\right) \\\\ &=\dfrac{1}{v}-\dfrac{1}{10}\\\\ \Rightarrow \dfrac{1}{v}&=\dfrac{1}{15}+\dfrac{1}{10}\\\\ &=\dfrac{2+3}{30}\\\\ &=\dfrac{1}{6}\\\\ \Rightarrow v&=6\\\\ m&=-\dfrac{v}{u}\\\\ &=-\dfrac{6}{-10}\\\\ &=0.6\end{aligned}$$ The image is formed 6 cm behind the mirror Image is erect, virtual & diminished.
13. The magnification produced by a plane mirror is +1. What does this mean?
In other words, a plane mirror always forms a virtual, erect image which is exactly the same size as the object. When a plane mirror produces a magnification of +1, it means:
- The image formed is upright (erect) due to the positive sign.
- The image is exactly the same size as the object (magnitude is 1).
- The image is virtual and formed as far behind the mirror as the object is in front.
14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of
radius of curvature 30 cm.
Find the position of the image, its nature and size.
Solution
Size of object \((h)\) = 5cm
Position of object \((u)\) = -20cm
Radius of Curvature of Convex mirror \((R)\) = 30cm
\(\therefore \) focal length of the convex mirror \(f=\frac{R}{2}=\frac{30}2{} = 15\ cm\)
$$\begin{aligned}\dfrac{1}{f}&=\dfrac{1}{v}+\dfrac{1}{u}\\\\
\dfrac{1}{15}&=\dfrac{1}{v}+\left( \dfrac{1}{-20}\right) \\\\
\dfrac{1}{15}&=\dfrac{1}{v}-\dfrac{1}{20}\\\\
\dfrac{1}{v}&=\dfrac{1}{15}+\dfrac{1}{20}\\\\
&=\dfrac{4+3}{60}\\\\
&=\dfrac{7}{60}\\\\
\Rightarrow v&=\dfrac{60}{7}cm\\\\
&=8.57\end{aligned}$$
Magnification \((m)\)
$$\begin{aligned}&m=-\dfrac{v}{u}\\\\
&=-\dfrac{\left( 60/7\right) }{-20}\\\\
&=\dfrac{60}{20\times 7}\\\\
&=\dfrac{3}{7}=0.43\\\\
m&=\dfrac{h'}{h}\\\\
h'&=m\cdot h\quad (h'\text{ height of Image})\\\\
&=\dfrac{3}{7}\times 5\\\\
&=\dfrac{15}{7}\\\\
&=2.14\ cm\end{aligned}$$
Image Position 8.57cm behind mirror Height = 2.14cm
Nature: Virtual, Erect, diminished
Q15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp, focused image can be obtained? Find the size and the nature of the image.
Solution
Size of the object \((h)\) = 7 cm
Postion of the object \((u)\) =-27cm
Focal length of the Mirror \((f)\) =-18cm
$$\begin{aligned}\dfrac{1}{f}&=\dfrac{1}{v}+\dfrac{1}{u}\\\\
\dfrac{1}{v}&=\dfrac{1}{t}-\dfrac{1}{u}\\\\
&=\dfrac{1}{-18}-\left( \dfrac{1}{-27}\right) \\\\
&=-\dfrac{1}{18}+\dfrac{1}{27}\\\\
&=\dfrac{-3+2}{54}\\\\
&=\dfrac{-1}{54}\\\\
\Rightarrow v&=-54\ cm\end{aligned}$$
magnification (m):
$$\begin{aligned}m&=-\dfrac{v}{u}\\
&=-\left( \dfrac{-54}{-27}\right) \\
&=-2\end{aligned}$$
Image Height h'
$$\begin{aligned}m&=\dfrac{h^{1}}{h}\\
-2&=\dfrac{h^{1}}{7}\\
\Rightarrow h^{1}&=-14\ cm\end{aligned}$$
Screen Position 54 em in front of Mirror Image size 14cm
Image Nature Real, Inverted, magnified
Q16. Find the focal length of a lens of power 2.0 D. What type of lens is this?
Solution
Power \(P\) of lens =-2D
Let focal Length of Lens =\(f\)
\[\begin{aligned}
P&=\frac{1}{f}\\\\-2&=\dfrac{1}{f}\\\\\Rightarrow f&=-\frac{1}{2}\ m\\\\&=-50\ cm
\end{aligned}\]
Focal Length is -ve, therefore the lens is a Concave lens
Q17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Solution Power = + 1. 5D $$\begin{aligned}P&=\dfrac{1}{f}\\\\ 1.5&=\dfrac{1}{f}\\\\ f&=\dfrac{1}{1.5}\\\\ &=\dfrac{10}{15}\\\\ &=\dfrac{2}{3}\\\\ &=0.67\ m\\ f&=67\ cm\end{aligned}$$ Since the focal length is +ve, therefore lens is a Converging/Convex Lens
