Q: A concave lens of focal length 15cm forms an image 10 cm from the lens . How far is the object placed from the lens? Draw the ray diagram.

Given: \$f=-15\$ \$v=-10\$From mirror formula, \$\cfrac{1}{v}-\cfrac{1}{u}=\cfrac{1}{f}\$ \$\cfrac{1}{-10}-\cfrac{1}{u}=\cfrac{1}{-15}\$ \$\cfrac{1}{u}=\cfrac{1}{15}-\cfrac{1}{10}\$ \$u=-30\ cm\$

Q: 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.

Given: \$f=15\$ \$u=-10\$From mirror formula, \$\cfrac{1}{u}+\cfrac{1}{v}=\cfrac{1}{f}\$ \$\cfrac{1}{-10}+\cfrac{1}{v}=\cfrac{1}{15}\$ \$v=6\ cm\$Image is formed 6 cm to the right of the mirror. Image formed is virtual and erect.

Q: No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

\$\textbf{Explanation:}\$ \$\bullet\$A virtual picture is formed by a plane mirror or a convex mirror, regardless of the object's position. \$\textbf{Option D is correct.}\$

Q: A doctor has prescribed a corrective lens of power \$+1.5D\$. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Power of lens \$p=+1.5D\$\$p=\dfrac { 1 }{ f } \$\$1.5=\dfrac { 1 }{ f } \$ \$f=\dfrac { 1 }{ 1.5 } m\$ \$f=\dfrac { 10 }{ 15 } m\$ \$f=\dfrac { 2 }{ 3 } m = 0.67m\$Focal length of the lens is \$+0.67m\$.The prescribed lens is converging in nature.

Q: 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.

Object distance should be less than the focal length for the formation of an erect image. Hence, the range of distance of object from the mirror should be less than 15 cm i.e. from 0 to 15 cm in the front of mirror from the pole. The nature of image so formed will be virtual and erect. The size of the image will be larger than object. Ray diagram is attached as image to the question.

Q: An object of size \$7.0 cm\$ is placed at \$27cm\$ 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 nature of the image.

Given:size of object, \$o=+7 cm\$distance of object ,\$u =-27 cm\$focal length of concave mirror, \$f=-18 cm\$let us take size of image=\$I\$so, mirror formula is \$\dfrac{1}{v} +\dfrac{1}{u} =\dfrac{1}{f}\$so, putting values of u and v,\$\dfrac{1}{v} =\dfrac{1}{-18} -\dfrac{1}{-27}\$\$\dfrac{1}{v}=\dfrac{1}{-54}\$\$v=-54cm\$so, image is formed on object side only.magnification,\$m=\dfrac{-v}{u} =\dfrac{I}{o}\$\$m=-\dfrac{-54}{-27} =\dfrac{I}{7}\$\$I=-14 cm\$so, image is double in size to that of object.Nature: real, inverted and magnified image.

Q: An object \$5 cm\$ is placed at a distance of \$20cm\$ in front of a convex mirror of radius of curvature \$30cm\$. Find the position of the image, its nature and size.

Formula:\$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\$Given,\$\dfrac{1}{v}+\dfrac{1}{(-20)}=\dfrac{1}{15}\$\$\dfrac{1}{v}=\dfrac{7}{60}\Rightarrow v=\dfrac{60}{7}\$\$v=8.57\approx 8.6cm\$The image is formed behind the mirror at a distance of \$8.6\ cm\$ \$m=\dfrac{h_2}{h_1}=-\dfrac{v}{u}\$\$\rightarrow \dfrac{h_2}{5}=-\dfrac{8.57}{20}\$\$\therefore h_2=2.175\ cm\$Thus the image is \$2.175\ cm\$ high, virtual and erect.Hence, this is the answer.

Q: Find the focal length of a lens of power (-2.0)D. What type of lens is this?

Given, lens of power \$P= (-2.0)D\$Power, \$P=\dfrac{1}{f}\$ \$f=\dfrac{1}{p}\$ \$=\dfrac{1}{-2}\$ \$f=-0.5\ m\$Since focal length is negative, it is concave lens.

Question 1

A concave lens of focal length 15cm forms an image 10 cm from the lens . How far is the object placed from the lens? Draw the ray diagram.

Accepted Answer

Given: $f=-15$            $v=-10$From mirror formula, $\cfrac{1}{v}-\cfrac{1}{u}=\cfrac{1}{f}$                                    $\cfrac{1}{-10}-\cfrac{1}{u}=\cfrac{1}{-15}$                                    $\cfrac{1}{u}=\cfrac{1}{15}-\cfrac{1}{10}$                                    $u=-30\ cm$

Question 2

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.

Accepted Answer

Given: $f=15$            $u=-10$From mirror formula, $\cfrac{1}{u}+\cfrac{1}{v}=\cfrac{1}{f}$                                   $\cfrac{1}{-10}+\cfrac{1}{v}=\cfrac{1}{15}$                                   $v=6\ cm$Image is formed 6 cm to the right of the mirror. Image formed is virtual and erect.

Question 3

No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

Accepted Answer

$\textbf{Explanation:}$ $\bullet$A virtual picture is formed by a plane mirror or a convex mirror, regardless of the object's position. $\textbf{Option D is correct.}$

Question 4

One half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Explain your observations.

Accepted Answer

When one-half of a convex lens is covered with a black paper, this lens produces a complete image of the object. To prove it we perform an experiment as given below: Take a convex lens and cover half part of it by using a black paper. Place it vertically in a stand. On one side of it place a burning candle. On opposite side of the lens fix a white screen. Adjust the position of candle or screen till clear image of burning candle is formed on the screen. We observe that the image is a complete image of the object (burning candle). From the experimental observations, we find that image formation does not depent upon the size of a lens. A smaller lens can also form complete image of an object placed in front of it. However, brightness of the image decreases when some part of lens is blocked. It is because now lesser number of rays pass through the lens.

Question 5

A doctor has prescribed a corrective lens of power $+1.5D$. Find the focal length of the lens. Is the prescribed lens diverging or converging?

Accepted Answer

Power of lens $p=+1.5D$$p=\dfrac { 1 }{ f } $$1.5=\dfrac { 1 }{ f } $ $f=\dfrac { 1 }{ 1.5 } m$ $f=\dfrac { 10 }{ 15 } m$ $f=\dfrac { 2 }{ 3 } m = 0.67m$Focal length of the lens is $+0.67m$.The prescribed lens is converging in nature.

Question 6

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.

Accepted Answer

Object distance should be&#10;less than the focal length for the formation of an erect image. Hence, the&#10;range of distance of object from the mirror should be less than 15 cm i.e. from&#10;0 to 15 cm in the front of mirror from the pole. The nature of image so formed&#10;will be virtual and erect. The size of the image will be larger than&#10;object. Ray diagram is attached as image to the question.

Question 7

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.

Accepted Answer

(a) Concave mirrorReason :Because to get powerful parallel beams of light.(b) Convex mirrorReason : Because these mirrors are fitted on the sides of the vehicle, enabling the driver to see traffic.(c) Concave MirrorReason : Because to get powerful parallel beams of light.

Question 8

An object of size $7.0 cm$ is placed at $27cm$ 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 nature of the image.

Accepted Answer

Given:size of object, $o=+7 cm$distance of object ,$u =-27 cm$focal length of concave mirror, $f=-18 cm$let us take size of image=$I$so, mirror formula is $\dfrac{1}{v} +\dfrac{1}{u} =\dfrac{1}{f}$so, putting values of u and v,$\dfrac{1}{v} =\dfrac{1}{-18} -\dfrac{1}{-27}$$\dfrac{1}{v}=\dfrac{1}{-54}$$v=-54cm$so, image is formed on object side only.magnification,$m=\dfrac{-v}{u} =\dfrac{I}{o}$$m=-\dfrac{-54}{-27} =\dfrac{I}{7}$$I=-14 cm$so, image is double in size to that of object.Nature: real, inverted and magnified image.

Question 9

An object $5 cm$ is placed at a distance of $20cm$ in front of a convex mirror of radius of curvature $30cm$. Find the position of the image, its nature and size.

Accepted Answer

Formula:$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$Given,$\dfrac{1}{v}+\dfrac{1}{(-20)}=\dfrac{1}{15}$$\dfrac{1}{v}=\dfrac{7}{60}\Rightarrow v=\dfrac{60}{7}$$v=8.57\approx 8.6cm$The image is formed behind the mirror at a distance of $8.6\ cm$ $m=\dfrac{h_2}{h_1}=-\dfrac{v}{u}$$\rightarrow \dfrac{h_2}{5}=-\dfrac{8.57}{20}$$\therefore h_2=2.175\ cm$Thus the image is $2.175\ cm$ high, virtual and erect.Hence, this is the answer.

Question 10

Find the focal length of a lens of power (-2.0)D. What type of lens is this?

Accepted Answer

Given,  lens of power $P= (-2.0)D$Power, $P=\dfrac{1}{f}$             $f=\dfrac{1}{p}$                $=\dfrac{1}{-2}$              $f=-0.5\ m$Since focal length is negative, it is concave lens.

Light - Reflection and Refraction

NCERT

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A concave lens of focal length 15cm forms an image 10 cm from the lens . How far is the object placed from the lens? Draw the ray diagram.

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.

No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

One half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Explain your observations.

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?

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.

An object of size $7.0 c m$ is placed at $27 c m$ in front of a concave mirror of focal length $18 c m$ . At what distance from the mirror should a screen be placed so that a sharp focused image can be obtained? Find the size and nature of the image.

An object $5 c m$ is placed at a distance of $20 c m$ in front of a convex mirror of radius of curvature $30 c m$ . Find the position of the image, its nature and size.

Find the focal length of a lens of power (-2.0)D. What type of lens is this?

Light - Reflection and Refraction

NCERT

Page 168

Page 171

Page 176

Page 184

Page 185

Page 186

A concave lens of focal length 15cm forms an image 10 cm from the lens . How far is the object placed from the lens? Draw the ray diagram.

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.

No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be

One half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Explain your observations.

A doctor has prescribed a corrective lens of power +1.5D. Find the focal length of the lens. Is the prescribed lens diverging or converging?

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.

An object of size 7.0cm is placed at 27cm in front of a concave mirror of focal length 18cm. At what distance from the mirror should a screen be placed so that a sharp focused image can be obtained? Find the size and nature of the image.

An object 5cm is placed at a distance of 20cm in front of a convex mirror of radius of curvature 30cm. Find the position of the image, its nature and size.

Find the focal length of a lens of power (-2.0)D. What type of lens is this?

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?

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.

An object of size $7.0 c m$ is placed at $27 c m$ in front of a concave mirror of focal length $18 c m$ . At what distance from the mirror should a screen be placed so that a sharp focused image can be obtained? Find the size and nature of the image.

An object $5 c m$ is placed at a distance of $20 c m$ in front of a convex mirror of radius of curvature $30 c m$ . Find the position of the image, its nature and size.