Question 1

Question 8The lens of power + 1.0 D is:convex of focal length 1.0 cmconvex of focal length 1.0 mconcave of focal length 1.0 cmconcave of focal length 1.0 m.

Accepted Answer

convex of focal length 1.0 mReason — As we know that,P=1fP = \dfrac{1}{f}P=f1​Where,P = power of lens andf = focal length of lens.Given,power = + 1.0 DSubstituting the values in the formula we get,+1.0 D=(1f) m⇒f=1.0 m+1.0 \text{ D} = \Big(\dfrac{1}{f}\Big)\text{ m} \[0.5em] \Rightarrow f = 1.0 \text{ m}+1.0 D=(f1​) m⇒f=1.0 mHence, focal length = 1.0 m

Question 2

Question 1(a) At what position a candle of length 3 cm be placed in front of a convex lens so that its image of length 6 cm be obtained on a screen placed at a distance 30 cm behind the lens?(b) What is the focal length of the lens in part (a)?

Accepted Answer

(a) As we know,The formula for magnification of a lens is —m=length of image (I)length of object (O)=vum = \dfrac{\text{length of image (I)}}{\text{length of object (O)}} = \dfrac{v}{u} \[0.5em]m=length of object (O)length of image (I)​=uv​where,v = image distanceu = object distance.Given,Height of a candle, O = 3 cmHeight of the image of candle, I = 6 cmImage distance, v = 30 cmSubstituting the values in the formula we get,63=30uu=30×36u=5×3⇒u=15\dfrac{6}{3} = \dfrac{30}{u} \[0.5em] u = \dfrac{30 \times 3}{6} \[0.5em] u = 5 \times 3 \[0.5em] \Rightarrow u = 1536​=u30​u=630×3​u=5×3⇒u=15Therefore, object distance is equal to 15 cm.(b) As we know, lens formula is —1v–1u=1f\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​–u1​=f1​Substituting the values in the formula we get,130–1−15=1f130+115=1f2+130=1f330=1ff=303⇒f=10\dfrac{1}{30} – \dfrac{1}{-15} = \dfrac{1}{f} \[0.5em] \dfrac{1}{30} + \dfrac{1}{15} = \dfrac{1}{f} \[0.5em] \dfrac{2 + 1}{30} = \dfrac{1}{f} \[0.5em] \dfrac{3}{30} = \dfrac{1}{f} \[0.5em] f = \dfrac{30}{3} \[0.5em] \Rightarrow f = 10301​–−151​=f1​301​+151​=f1​302+1​=f1​303​=f1​f=330​⇒f=10Hence, the focal length of the lens, f = 10 cm.

Question 3

Question 2A concave lens forms the image of an object kept at a distance 20 cm in front of it, at a distance 10 cm on the side of the object.(a) What is the nature of the image?(b) Find the focal length of the lens.

Accepted Answer

(a) As the lens used is a concave lens, so the image formed will be erect and diminished. It will be virtual in nature as the image is formed on the same side as the object.(b) As we know, the lens formula is —1v–1u=1f\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​–u1​=f1​Givenu = -20 cmv = – 10 cmSubstituting the values in the formula we get,1−10–1−20=1f−110+120=1f−2+120=1f−120=1f⇒f=−20cm\dfrac{1}{-10} – \dfrac{1}{-20} = \dfrac{1}{f} \[0.5em] -\dfrac{1}{10} + \dfrac{1}{20} = \dfrac{1}{f} \[0.5em] \dfrac{-2 + 1}{20} = \dfrac{1}{f} \[0.5em] -\dfrac{1}{20} = \dfrac{1}{f} \[0.5em] \Rightarrow f = -20 cm−101​–−201​=f1​−101​+201​=f1​20−2+1​=f1​−201​=f1​⇒f=−20cmTherefore, focal length of the lens = 20 cm (negative)

Question 4

Question 3The focal length of a convex lens is 25 cm. At what distance from the optical centre of the lens an object be placed to obtain a virtual image of twice the size ?

Accepted Answer

As we know,The formula for magnification of a lens is —m=vum = \dfrac{v}{u} \[0.5em]m=uv​Given,f = + 25 cmAs image is virtual and magnified, so m = + 2Substituting the values in the formula we get,+2=vu⇒v=2u+2 = \dfrac{v}{u} \[0.5em] \Rightarrow v = 2u \[0.5em]+2=uv​⇒v=2uTherefore, we get, v = 2uNow as we know, the lens formula is —1v–1u=1f\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​–u1​=f1​Substituting the values in the formula, we get,12u–1u=1251−22u=125−12u=1252u=−25⇒u=−12.5cm\dfrac{1}{2u} – \dfrac{1}{u} = \dfrac{1}{25} \[0.5em] \dfrac{1-2}{2u} = \dfrac{1}{25} \[0.5em] -\dfrac{1}{2u} = \dfrac{1}{25} \[0.5em] 2u = - 25 \[0.5em] \Rightarrow u = - 12.5 cm \[0.5em]2u1​–u1​=251​2u1−2​=251​−2u1​=251​2u=−25⇒u=−12.5cmTherefore, object should be placed at a distance of 12.5 cm infront of the lens.

Question 5

Question 4Where should an object be placed in front of a convex lens of focal length 0.12 m to obtain a real image of size three times the size of the object, on the screen?

Accepted Answer

(i) As we know,The formula for magnification of a lens is —m=vum = \dfrac{v}{u} \[0.5em]m=uv​Given,f = +0.12 mImage is real and magnified, m = -3Substituting the values in the formula we get,−3=vu⇒v=−3u-3 = \dfrac{v}{u} \[0.5em] \Rightarrow v = -3u \[0.5em]−3=uv​⇒v=−3uTherefore, we get, v = -3uNow as we know, the lens formula is —1v–1u=1f\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​–u1​=f1​Substituting the values in the formula, we get,1−3u–1u=10.12−1−3u3u=10.12−43u=10.12u=−4×0.123⇒u=−0.16m\dfrac{1}{-3u} – \dfrac{1}{u} = \dfrac{1}{0.12} \[0.5em] \dfrac{-1-3u}{3u} = \dfrac{1}{0.12} \[0.5em] -\dfrac{4}{3u} = \dfrac{1}{0.12}\[0.5em] u = - \dfrac{4 \times 0.12}{3} \[0.5em] \Rightarrow u = - 0.16 m \[0.5em]−3u1​–u1​=0.121​3u−1−3u​=0.121​−3u4​=0.121​u=−34×0.12​⇒u=−0.16mTherefore, object should be placed at a distance of 0.16 m infront of the lens.

Question 6

Question 5An illuminated object lies at a distance 1.0 m from a screen. A convex lens is used to form the image of the object on a screen placed at a distance of 75 cm from the lens.Find —(i) the focal length of lens, and(ii) the magnification.

Accepted Answer

As we know, the lens formula is —1v–1u=1f\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​–u1​=f1​Given,Object lies at a distance 1.0 m from a screen.v = 75 cmSo, u = - 25 cmSubstituting the values in the formula, we get,175–1−25=1f175+125=1f1+375=1f475=1f⇒f=754⇒f=18.75 cm\dfrac{1}{75} – \dfrac{1}{-25} = \dfrac{1}{f} \[0.5em] \dfrac{1}{75} + \dfrac{1}{25} = \dfrac{1}{f} \[0.5em] \dfrac{1+3}{75} = \dfrac{1}{f} \[0.5em] \dfrac{4}{75} = \dfrac{1}{f}\[0.5em] \Rightarrow f = \dfrac{75}{4} \[0.5em] \Rightarrow f = \text{18.75 cm} \[0.5em]751​–−251​=f1​751​+251​=f1​751+3​=f1​754​=f1​⇒f=475​⇒f=18.75 cmTherefore, focal length of the lens is 18.75cm.ii) As we know,the formula for magnification of a lens is —m=vum = \dfrac{v}{u} \[0.5em]m=uv​Given,v = 75 cmu = - 25 cmSubstituting the values in the formula, we get,m=75−25m=−3m = \dfrac{75}{-25} \[0.5em] m = -3 \[0.5em]m=−2575​m=−3Therefore, the magnification is -3.

Question 7

Question 6A lens forms the image of an object placed at a distance 15 cm from it, at a distance 60 cm in front of it.Find —(i) the focal length,(ii) the magnification, and(iii) the nature of image.

Accepted Answer

(i) As we know, the lens formula is —1v–1u=1f\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​–u1​=f1​Given,v = -60 cmSo, u = -15 cmSubstituting the values in the formula, we get,1−60–1−15=1f−160+115=1f115−160=1f4−160=1f360=1f⇒f=603⇒f=20cm\dfrac{1}{-60} – \dfrac{1}{-15} = \dfrac{1}{f} \[0.5em] - \dfrac{1}{60} + \dfrac{1}{15} = \dfrac{1}{f} \[0.5em] \dfrac{1}{15} - \dfrac{1}{60} = \dfrac{1}{f} \[0.5em] \dfrac{4-1}{60} = \dfrac{1}{f} \[0.5em] \dfrac{3}{60} = \dfrac{1}{f}\[0.5em] \Rightarrow f = \dfrac{60}{3} \[0.5em] \Rightarrow f = 20cm \[0.5em]−601​–−151​=f1​−601​+151​=f1​151​−601​=f1​604−1​=f1​603​=f1​⇒f=360​⇒f=20cmTherefore, focal length of the lens is 20 cm.ii) As we know,the formula for magnification of a lens is —m=vum = \dfrac{v}{u} \[0.5em]m=uv​ Given,v = -60 cmu = -15 cmSubstituting the values in the formula, we get,m=−60−15m=4m = \dfrac{- 60}{- 15} \[0.5em] m = 4 \[0.5em]m=−15−60​m=4Therefore, the magnification is 4.(iii) The nature of the image is erect, virtual and magnified.

Question 8

Question 7A lens forms the image of an object placed at a distance of 45 cm from it on a screen placed at a distance 90 cm on the other side of it.(a) Name the kind of lens.(b) Find : (i) the focal length of lens, and (ii) the magnification of the image.

Accepted Answer

(a) As the image is formed on the other side of the lens, so the image is real. Hence, the lens is convex.(b) (i) As we know, the lens formula is —1v–1u=1f\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​–u1​=f1​Given,u = – 45 cmv = + 90 cmSubstituting the values in the formula, we get,190–1−45=1f190+145=1f1+290=1f390=1f130=1f⇒f=30cm\dfrac{1}{90} – \dfrac{1}{-45} = \dfrac{1}{f} \[0.5em] \dfrac{1}{90} + \dfrac{1}{45} = \dfrac{1}{f} \[0.5em] \dfrac{1+2}{90} = \dfrac{1}{f} \[0.5em] \dfrac{3}{90} = \dfrac{1}{f} \[0.5em] \dfrac{1}{30} = \dfrac{1}{f}\[0.5em] \Rightarrow f = 30cm \[0.5em]901​–−451​=f1​901​+451​=f1​901+2​=f1​903​=f1​301​=f1​⇒f=30cmTherefore, focal length of the lens is 30 cm.ii) As we know,the formula for magnification of a lens is —m=vum = \dfrac{v}{u} \[0.5em]m=uv​ Given,u = –45 cmv = +90 cmSubstituting the values in the formula, we get,m=90−45m=−2m = \dfrac{90}{-45} \[0.5em] m = -2 \[0.5em]m=−4590​m=−2Therefore, the magnification is -2.

Question 9

Question 9A concave lens forms an erect image of 1/3rd the size of the object which is placed at a distance 30 cm in front of the lens. Find:(a) The position of the image, and(b) The focal length of the lens.

Accepted Answer

(a) As we know,the formula for magnification of a lens is —m=vu(Equation 1)m = \dfrac{v}{u} \qquad \bold{\text{(Equation 1)}}m=uv​(Equation 1)Given,u = -30 cmandm=IO=13(Equation 2)m = \dfrac{I}{O} = \dfrac{1}{3} \qquad \bold{\text{(Equation 2)}}m=OI​=31​(Equation 2)Substituting the values of u and Equation 2 in Equation 1, we get,v−30=13v=−10\dfrac{v}{-30} = \dfrac{1}{3} \[0.5em] v = -10 \[0.5em]−30v​=31​v=−10Therefore, the image is formed at 10 cm infront of the lens.(b) As we know, the lens formula is —1v–1u=1f\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​–u1​=f1​Given,u = -30 cmv = -10 cmThe lens used is concave in nature.Substituting the values in the formula, we get,1−10–1−30=1f−110+130=−1f−3+130=−1f−230=1f−115=1f⇒f=−15cm\dfrac{1}{-10} – \dfrac{1}{-30} = \dfrac{1}{f} \[0.5em] -\dfrac{1}{10} + \dfrac{1}{30} = - \dfrac{1}{f} \[0.5em] \dfrac{-3+1}{30} = - \dfrac{1}{f} \[0.5em] -\dfrac{2}{30} = \dfrac{1}{f} \[0.5em] -\dfrac{1}{15} = \dfrac{1}{f} \[0.5em] \Rightarrow f = -15 cm \[0.5em]−101​–−301​=f1​−101​+301​=−f1​30−3+1​=−f1​−302​=f1​−151​=f1​⇒f=−15cmTherefore, the focal length is 15 cm (negative)

Question 10

Question 10The power of a lens is +2.0 D. Find its focal length and state what kind of lens it is?

Accepted Answer

As we know,Power=1focal length\text{Power} = \dfrac{1}{\text{focal length}}Power=focal length1​Given,Power of the lens = +2.0DAs the given power is positive,Therefore, we can say that the lens used is convex in nature.Substituting the value of power in formula we get,2.0=1focal length⇒focal length=12⇒focal length=0.5m⇒focal length=50cm\text{2.0} = \dfrac{1}{\text{focal length}} \[0.5em] \Rightarrow \text{focal length} = \dfrac{1}{2} \[0.5em] \Rightarrow \text{focal length} = \text{0.5m} \[0.5em] \Rightarrow \text{focal length} = \text{50cm} \[0.5em]2.0=focal length1​⇒focal length=21​⇒focal length=0.5m⇒focal length=50cmTherefore, the focal length is 50 cm and the lens used is convex in nature.

Question 11

Question 11Express the power (with sign) of a concave lens of focal length 20 cm.

Accepted Answer

As we know,Power=1focal length\text{Power} = \dfrac{1}{\text{focal length}}Power=focal length1​Given,focal length = 20 cm = 0.2mSubstituting the value of focal length in formula we get,Power=10.2 m⇒Power=5D\text{Power} = \dfrac{1}{\text{0.2 m}} \[0.5em] \Rightarrow \text{Power} = 5 D \[0.5em]Power=0.2 m1​⇒Power=5DAs the lens used is concave in nature so power is negative.Hence, Power = – 5D

Question 12

Question 12The focal length of a convex lens is 25 cm. Express its power with sign.

Accepted Answer

As we know,Power=1focal length\text{Power} = \dfrac{1}{\text{focal length}}Power=focal length1​Given,focal length = 25 cm = 0.25mSubstituting the value of focal length in formula we get,Power=10.25 m⇒Power=4D\text{Power} = \dfrac{1}{\text{0.25 m}} \[0.5em] \Rightarrow \text{Power} = 4 D \[0.5em]Power=0.25 m1​⇒Power=4DAs the lens used is convex in nature so power is positive.Hence, P = +4D

Question 13

Question 13The power of a lens is -2.0 D. Find its focal length and its kind.

Accepted Answer

As we know,Power=1focal length\text{Power} = \dfrac{1}{\text{focal length}}Power=focal length1​Given,Power of the lens = -2.0 DAs 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,−2.0=1focal length⇒focal length=−12⇒focal length=−0.5m⇒focal length=−50cm-\text{2.0} = \dfrac{1}{\text{focal length}} \[0.5em] \Rightarrow \text{focal length} = -\dfrac{1}{2} \[0.5em] \Rightarrow \text{focal length} = -\text{0.5m} \[0.5em] \Rightarrow \text{focal length} = -\text{50cm} \[0.5em]−2.0=focal length1​⇒focal length=−21​⇒focal length=−0.5m⇒focal length=−50cmTherefore, the focal length is 50 cm and the lens used is concave in nature.

Question 14

Question 14The magnification by a lens is -3. Name the lens and state how are u and v related?

Accepted Answer

The magnification (m) of the lens is given as -3, which implies that the image is real and inverted because m is negative for a real and inverted image.The image is enlarged as the magnitude of magnification is greater than 1.As the image formed is real, inverted and magnified so the lens used is convex in nature.As we know, the relation between u and v is as shown below —m=vum = \dfrac{v}{u} \[0.5em]m=uv​Given,m = -3Substituting the values in the formula−3=vu⇒v=−3u-3 = \dfrac{v}{u} \[0.5em] \Rightarrow v = -3u \[0.5em]−3=uv​⇒v=−3uHence, the lens is convex in nature and the image distance v = -3u.

Question 15

Question 15The magnification by a lens is +0.5. Name the lens and state how are u and v related?

Accepted Answer

As the magnification (m) of the lens is given as +0.5, which implies that the image is virtual and erect because m is positive for a virtual and erect image.The image is smaller in size as the magnitude of magnification is smaller than 1.As the image formed is virtual, erect and smaller in size so the lens used is concave in nature.As we know, the relation between u and v is as shown below —m=vum = \dfrac{v}{u} \[0.5em]m=uv​Given,magnification by a lens is +0.5Substituting the values in the formula0.5=vu⇒v=0.5u0.5 = \dfrac{v}{u} \[0.5em] \Rightarrow v = 0.5u \[0.5em]0.5=uv​⇒v=0.5uThe lens is concave in nature and the image distance v = 0.5u.

Question 16

Question 16A photographer needs a lens with focal length of 0.2 m for close up shots. What is the power of the lens required ? How does a lens with higher power affect the focussing capability of a camera ?

Accepted Answer

Given,Focal length = f = 0.2mSo,P=1focal length (in metres)=10.2=102=5 D\text{P} = \dfrac{1}{\text{focal length (in metres)}} \[1em] = \dfrac{1}{0.2} \[1em] = \dfrac{10}{2} \[1em] = 5 \space \text{D}P=focal length (in metres)1​=0.21​=210​=5 DA lens with higher power has a shorter focal length, allowing it to focus on nearby objects more effectively. This makes it ideal for close-up shots.

Question 17

Question 17A concave lens has a focal length of 30 cm. Find the position and magnification (m) of the image for an object placed in front of it at a distance of 30 cm. State whether the image is real or virtual ?

Accepted Answer

As we know, the lens formula is —1v–1u=1f\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​–u1​=f1​Given,f = - 30 cmu = - 30 cmSubstituting the values in the formula, we get,1v–1−30=1−301v+130=−1301v=−130−1301v=−1−1301v=−2301v=−115⇒v=−15\dfrac{1}{v} – \dfrac{1}{-30} = \dfrac{1}{-30} \[0.5em] \dfrac{1}{v} + \dfrac{1}{30} = - \dfrac{1}{30} \[0.5em] \dfrac{1}{v} = -\dfrac{1}{30} - \dfrac{1}{30} \[0.5em] \dfrac{1}{v} = \dfrac{-1-1}{30} \[0.5em] \dfrac{1}{v} = -\dfrac{2}{30} \[0.5em] \dfrac{1}{v} = -\dfrac{1}{15} \[0.5em] \Rightarrow v = -15v1​–−301​=−301​v1​+301​=−301​v1​=−301​−301​v1​=30−1−1​v1​=−302​v1​=−151​⇒v=−15Therefore, the image in formed 15 cm in front of the lens.ii) As we know,the formula for magnification of a lens is —m=vum = \dfrac{v}{u} \[0.5em]m=uv​ Given,u = - 30 cmv = - 15 cmSubstituting the values in the formula, we get,m=−15−30m=0.5m = \dfrac{-15}{-30} \[0.5em] m = 0.5 \[0.5em]m=−30−15​m=0.5Therefore, the magnification is +0.5The image is virtual as the magnification is positive.

Question 18

Question 18Find the position and magnification of the image of an object placed at distance of 8.0 cm in front of a convex lens of focal length 10.0 cm. Is the image erect or inverted?

Accepted Answer

i) As the image is formed on the other side of the lens, so the image is real. Hence, the lens is convex.As we know, the lens formula is —1v–1u=1f\dfrac{1}{v} – \dfrac{1}{u} = \dfrac{1}{f} \[0.5em]v1​–u1​=f1​Given,f = 10 cmu = - 8 cmSubstituting the values in the formula, we get,1v–1−8=1101v+18=1101v=110−181v=4−5401v=−140⇒v=−40cm\dfrac{1}{v} – \dfrac{1}{-8} = \dfrac{1}{10} \[0.5em] \dfrac{1}{v} + \dfrac{1}{8} = \dfrac{1}{10} \[0.5em] \dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{8} \[0.5em] \dfrac{1}{v} = \dfrac{4-5}{40} \[0.5em] \dfrac{1}{v} = -\dfrac{1}{40} \[0.5em] \Rightarrow v = - 40 cm \[0.5em]v1​–−81​=101​v1​+81​=101​v1​=101​−81​v1​=404−5​v1​=−401​⇒v=−40cmTherefore, the image is formed 40 cm in front of the lens.ii) As we know,the formula for magnification of a lens is —m=vum = \dfrac{v}{u} \[0.5em]m=uv​ Given,u = –8 cmv = -40 cmSubstituting the values in the formula, we get,m=−40−8m=5m = \dfrac{-40}{-8} \[0.5em] m = 5 \[0.5em]m=−8−40​m=5Therefore, the magnification is +5.0As the magnification is positive so the image is erect.

Refraction Through a Lens

Exercise 5C Multiple Choice Type

Exercise 5C Numericals