NCERT Solutions Class 10 Science Chapter 10

Q.1 Which one of the following materials cannot be used to make a lens?

(a) Water
(b) Glass
(c) Plastic
(d) Clay

Ans-

The correct option is (d).
Explanation: The light can pass through a lens. As clay does not allow light to pass through it hence, it cannot be used to make a lens.

Q.2 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 principle 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 principle focus.

Ans-

The correct option is (d).
Explanation: Concave mirror forms a virtual, erect and enlarged image of an object when the object is placed between the pole and focus of the mirror.

Q.3 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 principle focus of the lens
(b) At twice the focal length
(c) At infinity
(d) Between the optical centre of the lens and its principle focus.

Ans-

The correct option is (b).

Explanation: On placing the object at a distance twice of the focal length (at centre of curvature) in front of a convex lens, the image is formed at the centre of curvature on the other side of the lens. This image is real, inverted, and of the same size as the object.

Q.4 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 and the lens is concave.

Ans-

The correct option is (a).
Explanation: As per sign convention, the focal length of a concave mirror and a concave lens are considered as negative.

Q.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.

Ans-

The correct option is (d).
Explanation: A plane mirror always forms a virtual and erect image of same size as that of the object. In the same way, a convex mirror forms a virtual and erect image of smaller size of the object placed in front of it.

Q.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 of 50 cm
(c) A convex lens of focal length of 5 cm
(d) A concave lens of focal length 5 cm

Ans-

The correct option is (c).
Explanation: When a convex lens is placed between the radius of curvature and focal length, it forms magnified image of the given object. Moreover, magnification is more for convex lenses having shorter focal length.

Q.7 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 distance 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.

Ans-

A concave mirror always forms an erect image when object is placed between pole (P) and the principal focus (F). Therefore, the object should be placed anywhere between the pole and the focus of the concave mirror. The image formed in this case will be virtual, erect, and magnified in nature.

Q.8 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 reason.

Ans-

(a) Concave mirror
(b) Convex mirror
(c) Concave mirror.

Reasons:
(a) In the headlights of cars, concave mirror is used as it produces parallel beam of light when the light source (bulb) is at the principal focus of the concave mirror.
(b) A convex mirror forms a virtual, erect, and diminished image of the objects placed in front of. Hence, a driver can see most of the traffic behind him.
(c) A concave mirror can converge the light incident on it at a single point known as principal focus. Therefore, it can be used to produce a large amount of heat at that point.

Q.9 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.

Ans-

The convex lens will produce a compete image of the object kept in front of it even half of the lens is covered with black paper.

i. When the upper half of the lens is covered:

The rays of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object.

ii. When the lower half of the lens is covered:

The rays of light coming from the object are refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object.

Q.10 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.

Ans-

\begin{array}{l}\text{Given},\text{object distance (u) = -25 cm}\\ {\text{object height (h}}_{\text{o}}\text{)=5 cm}\\ \text{focal length (f) = +10 cm}\\ \text{As per lens formula,}\\ \frac{\text{1}}{\text{v}}-\frac{\text{1}}{\text{u}}=\frac{\text{1}}{\text{f}}\\ \Rightarrow \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\text{+}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{10}}-\frac{\text{1}}{\text{25}}\text{=}\frac{\text{15}}{\text{250}}\\ \Rightarrow \text{v}\text{=}\frac{\text{250}}{\text{15}}\text{=}\text{16}\text{.66 cm}\\ \text{As image distance is positive hence, the image is}\\ \text{formed at the other side of the lens}\text{.}\\ \text{Magnification (m) =}\frac{image\text{distance}}{object\text{distance}}\text{=}-\frac{\text{v}}{\text{u}}\text{=}-\frac{\text{16}\text{.66}}{\text{25}}\\ \text{=}-\text{0}\text{.66}\\ \text{Here, negative sign shows that the image is real and}\\ \text{formed behind the lens}\text{.}\\ \text{Again, Magnification (m)}\text{=}\frac{\text{image height}}{\text{object height}}\text{=}\frac{{\text{h}}_{\text{i}}}{{\text{h}}_{\text{o}}}\text{=}\frac{{\text{h}}_{\text{i}}}{\text{5}}\\ {\text{or, h}}_{\text{i}}\text{=}\text{m}\text{×}{\text{h}}_{\text{o}}\text{=}-\text{0}\text{.66}\text{×}\text{5}\text{=}-\text{3}\text{.3 cm}\\ \text{Hence, the image formed is inverted}\text{.}\end{array}

Q.11 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.

Ans-

\begin{array}{l}\text{Given, focal length of concave lens}\left(\text{f}\right)\text{=}-\text{15 cm}\\ \text{image distance}\left(\text{v}\right)\text{=}-\text{10 cm}\\ \text{On using lens formula,}\\ \frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \Rightarrow \frac{\text{1}}{\text{v}}\text{–}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\text{=}-\frac{\text{1}}{\text{10 cm}}-\frac{\text{1}}{\left(\text{-15 cm}\right)}\text{=}-\frac{\text{5}}{\text{150 cm}}\\ \Rightarrow \text{u =}-\text{30 cm}\\ \text{Here, negative object distance shows that the object is placed 30 cm in front of the lens}\text{.}\\ \end{array}

Q.12 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.

Ans-

\begin{array}{l}\text{Given, focal length}\left(\text{f}\right)\text{= +15 cm}\\ \text{object distance}\left(\text{u}\right)\text{=}-\text{10 cm}\\ \text{As per mirror formula,}\\ \frac{\text{1}}{\text{v}}+\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \Rightarrow \frac{\text{1}}{\text{v}}=\frac{\text{1}}{\text{f}}-\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{15 cm}}-\frac{\text{1}}{\left(\text{-10 cm}\right)}\text{=}\frac{\text{25}}{\text{150 cm}}\\ \text{v = 6 cm}\\ \text{As}\left(\text{v}\right)\text{is positive hence, image is formed behind the mirror}\text{.}\\ \text{Magnification}\left(\text{m}\right)\text{=}\frac{\text{image distance}}{\text{object distance}}\text{=}-\frac{\text{v}}{\text{4}}\text{=}-\frac{\text{6}}{\left(-\text{10}\right)}\text{=+0}\text{.6}\\ \text{Here, positive sign shows that the image formed is virtual and erect}\text{.}\end{array}

Q.13 The magnification produced by a plane mirror is +1. What does this mean?

Ans-

\begin{array}{l}For\text{a mirror, magnification (m) is given as,}\\ \text{m =}\frac{h_i}{h_o}\\ Where,{\text{h}}_{\text{i}}={\text{image height , h}}_{\text{o}}=\text{object height}\\ \text{When magnification produced by a plane mirror is +1,}\\ \text{the image formed would be of the same size as that}\\ \text{of object}\text{. Also, the positive sign indicates that the}\\ \text{image formed is virtual and erect}\text{.}\end{array}

Q.4 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.

Ans-

\begin{array}{l}\text{Given,}\text{u}\text{=}\text{-20}\text{cm,}{\text{h}}_{\text{o}}\text{=}\text{5}\text{cm,}\text{R}\text{=}\text{30}\text{cm}\\ \text{focal}\text{length}\text{(f)}\text{=}\frac{\text{R}}{\text{2}}\text{=}\frac{\text{30 cm}}{\text{2 cm}}\text{=}\text{15}\text{cm}\\ \text{As}\text{per}\text{mirror}\text{formula,}\\ \frac{\text{1}}{\text{v}}\text{+}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{f}}\\ \Rightarrow \frac{\text{1}}{\text{v}}\text{=}\frac{\text{1}}{\text{f}}\text{–}\frac{\text{1}}{\text{u}}\text{=}\frac{\text{1}}{\text{15 cm}}\text{+}\frac{\text{1}}{\text{20 cm}}\text{=}\frac{\text{7}}{\text{60 cm}}\\ \text{v =}\text{8}\text{.57}\text{cm}\\ \text{As}\text{v}\text{is}\text{positive}\text{hence,}\text{image}\text{is}\text{formed}\text{behind}\text{the}\text{mirror}\text{.}\\ \text{Magnification}\text{(m)}\text{=}\text{–}\frac{\text{v}}{\text{u}}\text{=}\text{–}\frac{\text{8}\text{.57 cm}}{\text{(-20 cm)}}\text{=}\text{0}\text{.428}\\ \text{The}\text{positive}\text{value}\text{of}\text{magnification}\text{shows}\text{that}\text{the}\text{image}\text{formed}\text{is}\text{virtual}\text{.}\\ \text{Again,}\text{Magnification}\text{(m)}\text{=}\frac{{\text{h}}_{\text{i}}}{{\text{h}}_{\text{o}}}\\ \text{or,}{\text{h}}_{\text{i}}\text{=}{\text{m × h}}_{\text{o}}\text{=}\text{0}\text{.428}\text{×5}\text{cm}\text{=}\text{2}\text{.14}\text{cm}\\ \text{The}\text{positive}\text{value}\text{of}\text{image}\text{height}\text{indicates}\text{that}\text{the image}\text{formed}\text{is}\text{erect}\text{.}\\ \text{Therefore,}\text{the}\text{image}\text{formed}\text{is}\text{virtual,}\text{erect}\text{and}\text{smaller in}\text{size}\text{.}\end{array}

Q.15 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 focussed image can be obtained? Find the size and the nature of the image.

Ans-

$\begin{array}{l}\mathrm{Given},\text{u = -27 cm}\\ {\text{h}}_{\text{o}}=\text{7 cm}\\ \text{f = -18 cm}\\ \text{As per mirror formula,}\\ \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\\ \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{\mathrm{f}}-\frac{1}{\mathrm{u}}=\frac{1}{-18}-\frac{1}{(-27)}=-\frac{1}{54}\\ \text{v= -54 cm}\\ \mathrm{Hence},\text{the screen should be placed at a distance of}\\ \text{54 cm in front of the given mirror.}\\ \text{Magnification (m) =}-\frac{\mathrm{v}}{\mathrm{u}}=-\frac{(-54)}{(-27)}=-2\\ \mathrm{Here},\text{negative sign shows that the image formed is real.}\\ \mathrm{Again},\text{m =}\frac{{\mathrm{h}}_{\mathrm{i}}}{{\mathrm{h}}_{\mathrm{o}}}\\ \mathrm{or},{\text{h}}_{\text{i}}=\text{m}\times {\mathrm{h}}_{\mathrm{o}}=7\times (-2)=-14\mathrm{cm}\\ \mathrm{The}\text{negative value of image height shows that the image}\\ \text{formed is inverted.}\end{array}$

Q.16 Find the focal length of a lens of power – 2.0 D. What type of lens is this?

Ans-

$\begin{array}{l}\text{Given, power of lens, P = -2 D}\\ \text{P =}\frac{1}{\text{f}\left(\text{metres}\right)}\\ \therefore \text{f =}\frac{1}{2D}=–0.5\text{m}\\ \text{As a concave lens has a negative focal length.}\\ \text{Hence, this lens is a concave.}\end{array}$

Q.17 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?

Ans-

\begin{array}{l}\text{Given, power of lens P = 1}\text{.5 D}\\ \text{On using the formula for power}\\ \text{P =}\frac{1}{\text{f}\left(\text{metres}\right)}\\ \therefore \text{f =}\frac{1}{\mathrm{1.5\; D}}=0.66\text{m}\\ \text{As a convex lens has a positive focal length}\text{. Hence it is convex lens}\text{.}\end{array}