Question 1

Question 1Which one of the following materials cannot be used to make a lens?WaterGlassPlasticClay

Accepted Answer

ClayReason — Clay cannot be used to make a lens as it is opaque and will not allow light to pass through it.

Question 2

Question 2The 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?Between the principal focus and the centre of curvatureAt the centre of curvatureBeyond the centre of curvatureBetween the pole of the mirror and its principal focus.

Accepted Answer

Between the pole of the mirror and its principal focusReason — In order to get a virtual, erect and larger image of an object with the help of a concave mirror, the object should be placed between the pole of the mirror and its principal focus.

Question 3

Question 3Where should an object be placed in front of a convex lens to get a real image of the size of the object?At the principal focus of the lensAt twice the focal lengthAt infinityBetween the optical centre of the lens and its principal focus.

Accepted Answer

At twice the focal lengthReason — In order to get a real image of the same size as the object with the help of a convex lens, the object should be placed at twice the focal length.

Question 4

Question 4A spherical mirror and a thin spherical lens have a focal length of -15 cm. The mirror and the lens are likely to beboth concaveboth convexthe mirror is concave, and the lens is convexthe mirror is convex, but the lens is concave

Accepted Answer

both concaveReason — According to the sign convention, the focal length for both a concave mirror and a concave lens is negative.

Question 5

Question 5No matter how far you stand from a mirror, your image appears erect. The mirror is likely to beonly planeonly concaveonly convexeither plane or convex

Accepted Answer

either plane or convexReason — Both plane mirrors and concave mirrors form erect images irrespective of the position of the object. Concave mirror only forms erect image when the object is placed between it's pole and focus.

Question 6

Question 6Which of the following lenses would you prefer to use while reading small letters found in a dictionary?A convex lens of focal length 50 cmA concave lens of focal length 50 cmA convex lens of focal length 5 cmA concave lens of focal length 5 cm

Accepted Answer

A convex lens of focal length 5 cmReason — A convex lens gives a magnified image and shorter the focal length, more is the magnification, hence, a convex lens of focal length 5 cm will be preferred.

Question 7

Question 7We 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

Concave mirror gives a virtual and erect image only when the object is placed between its pole and focus.Hence, the range of distance of the object should be less than 15 cm from the mirror.Image is virtual.Image is larger than the object.Ray diagram to obtain an erect image using a concave mirror is shown below:

Question 8

Question 8Name the type of mirror used in the following situations.(a) Headlights of a car(b) Side/rear-view mirror of a vehicle(c) Solar furnaceSupport your answer with a reason.

Accepted Answer

(a) Concave mirrors because they produce powerful parallel beam of light when the light source is placed at its principal focus.(b) Convex Mirror because they always give an erect image with a wider field of view as they are curved outward, enabling the driver to view much larger area than would be possible with a plane mirror.(c) Concave Mirror because it concentrates the parallel rays of the sun at a principal focus.

Question 9

Question 9One-half of a convex lens is covered with black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.

Accepted Answer

Yes, it will produce a complete image of the object. However, the intensity or brightness of the image will be reduced.Experiment : To show that a half-covered convex lens can produce a complete image of the object.Apparatus needed : A convex lens, a candle, and two black sheets of paper.Procedure :Place a convex lens on the table. Light a candle with a match stick and place it near one side of the lens on the table (turn off the lights of the room in order to get dark).Take the sheet of paper and place it on the other side of the lens as the candle.Now focus the image formed on the sheet of paper by moving the sheet only.Observe the image carefully.Mark the positions of candle, lens and screen.Now, turn off the light. Take the convex lens and cover half of it with the second sheet of black paper.Mount the lens on the table in the original position and repeat the above experiment.Observations : A complete image of the object is obtained in both the cases. However, the image formed with half of the convex lens covered is less bright.Conclusion : The half covered lens will also produce a complete image of the object. However, the image formed will be less bright.

Question 10

Question 10An 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 nature of the image formed.

Accepted Answer

Given,Object height (ho) = 5 cmObject distance (u) = -25 cmfocal length (f) = 10 cmRay diagram is shown below:Image distance (v) = ?Image height (hi) = ?According to the Lens formula,1v\dfrac{1}{v}v1​ - 1u\dfrac{1}{u}u1​ = 1f\dfrac{1}{f}f1​Substituting the values we get,1v−1−25=1101v+125=110⇒1v=110−125⇒1v=5−250⇒1v=350⇒v=16.66 cm\dfrac{1}{v} - \dfrac{1}{-25} = \dfrac{1}{10} \[0.5em] \dfrac{1}{v} + \dfrac{1}{25} = \dfrac{1}{10} \[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 10} - \dfrac{1}{25} \[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{5-2}{50} \[0.5em] \Rightarrow \dfrac{1}{v} = \dfrac{3}{50} \[0.5em] \Rightarrow v = 16.66 \text{ cm}v1​−−251​=101​v1​+251​=101​⇒v1​=101​−251​⇒v1​=505−2​⇒v1​=503​⇒v=16.66 cmTherefore, the distance of the image is 16.66 cm on the opposite side of the lens.Magnification (m) = vu\dfrac{\text{v}}{\text{u}}uv​ = height of imageheight of object\dfrac{\text{height of image}}{\text{height of object}}height of objectheight of image​Substituting the values we get,16.66−25\dfrac{16.66}{-25}−2516.66​ = height of image5\dfrac{\text{height of image}}{5}5height of image​height of image = 16.66×5−25\dfrac{16.66 \times 5}{-25}−2516.66×5​ = -3.3Negative sign of the height of image means that an inverted image is formed.The image is reduced, real and inverted.

Question 11

Question 11A 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.

Accepted Answer

f = -15 cm [∵ focal length of concave lens is negative]v = -10 cm [∵ image distance of concave lens is negative]u = ?According to the lens formula,1v\dfrac{1}{v}v1​ - 1u\dfrac{1}{u}u1​ = 1f\dfrac{1}{f}f1​Substituting the values we get,1−10−1u=1−15⇒1u=1−10+115⇒1u=−3+230⇒1u=−130⇒u=−30 cm\dfrac{1}{- 10} - \dfrac{1}{u} = \dfrac{1}{- 15} \[0.5em] \Rightarrow \dfrac{1}{u} = \dfrac{1}{- 10} + \dfrac{1}{15} \[0.5em] \Rightarrow \dfrac{1}{u} = \dfrac{-3+2}{30} \[0.5em] \Rightarrow \dfrac{1}{u} = - \dfrac{1}{30} \[0.5em] \Rightarrow u = -30\text{ cm}−101​−u1​=−151​⇒u1​=−101​+151​⇒u1​=30−3+2​⇒u1​=−301​⇒u=−30 cmThe object is placed 30 cm in front of the concave lens.

Question 12

Question 12An 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 cmu = -10 cmv = ?According to the mirror formula,1v\dfrac{1}{v}v1​ + 1u\dfrac{1}{u}u1​ = 1f\dfrac{1}{f}f1​Substituting the values we get,1v+1−10=115⇒1v=110+115⇒1v=3+230⇒1v=530⇒1v=16⇒v=6 cm\dfrac{1}{v} + \dfrac{1}{-10} = \dfrac{1}{15} \[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 10} + \dfrac{1}{15} \[1em] \Rightarrow \dfrac{1}{v} = \dfrac{3+2}{30} \[1em] \Rightarrow \dfrac{1}{v} = \dfrac{5}{30} \[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{6} \[1em] \Rightarrow v = 6\text{ cm}v1​+−101​=151​⇒v1​=101​+151​⇒v1​=303+2​⇒v1​=305​⇒v1​=61​⇒v=6 cmTherefore, image is formed 6 cm behind the mirror. Image is virtual and erect.

Question 13

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

Accepted Answer

Magnification=−vu=height of imageheight of object\text{Magnification} = \dfrac{-v}{u} = \dfrac{\text{height of image}}{\text{height of object}} \[1em]Magnification=u−v​=height of objectheight of image​Given,+1=height of imageheight of object∴height of image=height of object+1 = \dfrac{\text{height of image}}{\text{height of object}} \[1em] ∴ \text{height of image} = \text{height of object}+1=height of objectheight of image​∴height of image=height of objectHence, it means that the image formed by a plane mirror will be virtual and erect and it will be of the same size as the object.

Question 14

Question 14An 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.

Accepted Answer

Given,Object distance (u) = -20 cmObject height (h) = 5 cmImage distance (v) = ?Image height (h) = ?Radius of curvature (R) = 30 cmWe know, R = 2f or f = R2\dfrac{R}{2}2R​Hence, f = 302\dfrac{30}{2}230​ = 15 cmAccording to the mirror formula,1v\dfrac{1}{v}v1​ + 1u\dfrac{1}{u}u1​ = 1f\dfrac{1}{f}f1​Substituting the values we get,1v+1−20=115⇒1v=120+115⇒1v=3+460⇒1v=760⇒v=607=8.57 cm\dfrac{1}{v} + \dfrac{1}{-20} = \dfrac{1}{15} \[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 20} + \dfrac{1}{15} \[1em] \Rightarrow \dfrac{1}{v} = \dfrac{3+4}{60} \[1em] \Rightarrow \dfrac{1}{v} = \dfrac{7}{60} \[1em] \Rightarrow v = \dfrac{60}{7} = 8.57\text{ cm}v1​+−201​=151​⇒v1​=201​+151​⇒v1​=603+4​⇒v1​=607​⇒v=760​=8.57 cmImage is virtual, erect and formed 8.57 cm behind the mirror.Magnification=−vu=height of imageheight of object−8.57−20=height of image5⇒height of image=−8.57×5−20⇒height of image=2.14 cm\text{Magnification} = \dfrac{-v}{u} = \dfrac{\text{height of image}}{\text{height of object}} \[0.5em] \dfrac{-8.57}{-20} = \dfrac{\text{height of image}}{5} \[0.5em] \Rightarrow \text{height of image} = \dfrac{-8.57 \times 5}{-20} \[0.5em] \Rightarrow \text{height of image} = 2.14 \text{ cm}Magnification=u−v​=height of objectheight of image​−20−8.57​=5height of image​⇒height of image=−20−8.57×5​⇒height of image=2.14 cmHeight of image is 2.14 cm

Question 15

Question 15An 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.

Accepted Answer

Given,Object distance (u) = -27 cm [∵ object is placed infront of mirror so object distance will be negative]Object height (h) = 7 cmImage distance (v) = ?Image height (h) = ?f = -18 cm [∵ focal length of concave mirror is negative by convention]According to the mirror formula,1v\dfrac{1}{v}v1​ + 1u\dfrac{1}{u}u1​ = 1f\dfrac{1}{f}f1​Substituting the values we get,1v+1−27=1−18⇒1v=127−118⇒1v=2−354⇒1v=−154⇒v=−54 cm\dfrac{1}{v} + \dfrac{1}{-27} = \dfrac{1}{-18} \[1em] \Rightarrow \dfrac{1}{v} = \dfrac{1}{ 27} - \dfrac{1}{18} \[1em] \Rightarrow \dfrac{1}{v} = \dfrac{2-3}{54} \[1em] \Rightarrow \dfrac{1}{v} = - \dfrac{1}{54} \[1em] \Rightarrow v = - 54\text{ cm}v1​+−271​=−181​⇒v1​=271​−181​⇒v1​=542−3​⇒v1​=−541​⇒v=−54 cm∴ The screen should be placed at a distance of 54 cm in front of the mirror on the object side.Magnification=−vu=height of imageheight of object−−54−27=height of image7⇒height of image=−54×727⇒height of image=−14 cm\text{Magnification} = \dfrac{-v}{u} = \dfrac{\text{height of image}}{\text{height of object}} \[1em] -\dfrac{-54}{-27} = \dfrac{\text{height of image}}{7} \[1em] \Rightarrow \text{height of image} = -\dfrac{54 \times 7}{27} \[1em] \Rightarrow \text{height of image} = -14 \text{ cm}Magnification=u−v​=height of objectheight of image​−−27−54​=7height of image​⇒height of image=−2754×7​⇒height of image=−14 cm∴ Height of image is 14 cm and image is real, inverted and magnified.

Question 16

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

Accepted Answer

Given,P = -2.0 Df = ?We know,Power of lens (P) = 1focal length\dfrac{1}{	ext{focal length}}focal length1​As the given power is negative hence, we can say that the lens used is concave in nature.Substituting the value of power in formula we get,-2 = 1f\dfrac{1}{	ext{f}}f1​f = - 12\dfrac{1}{	ext{2}}21​f = -0.5 mTherefore, focal length is -0.5 m and it is a concave lens.

Question 17

Question 17A 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?

Accepted Answer

Given,P = +1.5 Df = ?We know,Power of lens (P) = 1focal length\dfrac{1}{	ext{focal length}}focal length1​As the given power is positive hence, we can say that the lens used is convex (converging) in nature.Substituting the value of power in formula we get,+1.5 = 1f\dfrac{1}{	ext{f}}f1​f = 11.5\dfrac{1}{	ext{1.5}}1.51​f = 0.67 mTherefore, focal length is 0.66 m and it is a converging lens.

Question 18

Question 1Define the principal focus of a concave mirror.

Accepted Answer

The principal focus of a concave mirror is a point on the principal axis through which the light rays incident parallel to the principal axis, pass after reflection from the mirror.

Question 19

Question 2The radius of curvature of a spherical mirror is 20 cm. What is its focal length?

Accepted Answer

Given,Radius of curvature (R) = 20 cmFocal length (f) = ?We know, R = 2f or f = R2\dfrac{R}{2}2R​Hence, f = 202\dfrac{20}{2}220​ = 10 cm∴ Focal length of spherical mirror is 10 cm.

Question 20

Question 3Name the mirror that can give an erect and enlarged image of an object.

Accepted Answer

Concave Mirror

Light - Reflection and Refraction

Exercises

Intext Questions 1

Intext Questions 2

Intext Questions 3

Intext Questions 4