Q: Question 5For a ray reflected on a plane mirror, find the ratio of sin i/sin r

The ratio of sin isin r\dfrac{\text{sin i}}{\text{sin r}}sin rsin i = 1According to the laws of reflection angle of incidence = angle of reflection, hence sin i = sin r and we get the ratio as 1.

Q: Question 1The radius of curvature of a convex mirror is 40 cm. find it's focal length.

Focal length = 12\dfrac{1}{2}21 x Radius of curvatureGiven,R = 40 cmSubstituting the values in the formula above we get,Focal length=12×40Focal length=20 cm\text{Focal length} = \dfrac{1}{2} \times 40 \\[0.5em] \text{Focal length} = 20 \text{ cm}Focal length=21×40Focal length=20 cmHence, focal length of the convex mirror = 20 cm.

Q: Question 2The focal length of a concave mirror is 10 cm. Find it's radius of curvature.

Focal length = 12\dfrac{1}{2}21 x Radius of curvatureGiven,f = 10 cmSubstituting the values in the formula above we get,10=12×radius of curvature⇒radius of curvature=20 cm10 = \dfrac{1}{2} \times \text{radius of curvature} \\[0.5em] \Rightarrow \text{radius of curvature} = 20 \text{ cm}10=21×radius of curvature⇒radius of curvature=20 cmHence, radius of curvature of concave mirror = 20 cm.

Question 1

Question 5For a ray reflected on a plane mirror, find the ratio of sin i/sin r

Accepted Answer

The ratio of sin isin r\dfrac{	ext{sin i}}{	ext{sin r}}sin rsin i​ = 1According to the laws of reflection angle of incidence = angle of reflection, hence sin i = sin r and we get the ratio as 1.

Question 2

Question 1(xvi)The type of mirror used as a shaving mirror in daily life is :plane mirrorconvex mirrorconcave mirrorparabolic mirror

Accepted Answer

concave mirrorReason — When a concave mirror is held near the face (such that the face is between pole and focus of the mirror), it gives an upright and magnified image. Hence even tiny hair can be seen. For this concave mirror of large focal length and large aperture is used.

Question 3

Question 1(xvii)Which mirror always forms a diminished image for all positions of the object placed in front of it ?plane mirrorconvex mirrorconcave mirrorparabolic mirror

Accepted Answer

convex mirrorReason — A convex mirror always forms a diminished image for all positions of the object placed in front of it.

Question 4

Question 1(xviii)The figure shows the image of a clock as seen in the plane mirror. The correct time is :4:306:302:303:30

Accepted Answer

2:30Reason — As the mirror image of the clock shows that hour hand is pointing slightly past 9 and minute hand at 6 (i.e., 30 minutes) so the mirror image shows 9:30. Since a plane mirror always produces laterally inverted images so in this case the hour hand should actually point slightly past 2 and minute hand will suffer no inversion due to symmetry so the clock will show 2:30.

Question 5

Question 1(xix)We use a concave polished metallic surface as a reflector in a torch to obtain a parallel beam of light. The position of the bulb is :at the centre of curvature of the reflector.at the focus of the reflectorbetween the focus and centre of curvature of the reflectorcan be any of the above options

Accepted Answer

at the focus of the reflectorReason — In a torch, we want the light rays to come out as a parallel beam as this helps the light travel a long distance in a specific direction and a concave mirror reflects light rays parallel to the principal axis if the source is placed at its focus.So, if we want a parallel beam of light (which is the case in a torch), the bulb should be placed at the focus of the concave mirror.

Question 6

Question 1The radius of curvature of a convex mirror is 40 cm. find it's focal length.

Accepted Answer

Focal length = 12\dfrac{1}{2}21​ x Radius of curvatureGiven,R = 40 cmSubstituting the values in the formula above we get,Focal length=12×40Focal length=20 cm\text{Focal length} = \dfrac{1}{2} \times 40 \[0.5em] \text{Focal length} = 20 \text{ cm}Focal length=21​×40Focal length=20 cmHence, focal length of the convex mirror = 20 cm.

Question 7

Question 2The focal length of a concave mirror is 10 cm. Find it's radius of curvature.

Accepted Answer

Focal length = 12\dfrac{1}{2}21​ x Radius of curvatureGiven,f = 10 cmSubstituting the values in the formula above we get,10=12×radius of curvature⇒radius of curvature=20 cm10 = \dfrac{1}{2} \times \text{radius of curvature} \[0.5em] \Rightarrow \text{radius of curvature} = 20 \text{ cm}10=21​×radius of curvature⇒radius of curvature=20 cmHence, radius of curvature of concave mirror = 20 cm.

Question 8

Question 3An object of height 2 cm is placed at a distance 20 cm in front of a concave mirror of focal length 12 cm. Find the position, size and nature of the image.

Accepted Answer

Given,Object height (O) = 2 cmFocal length (f) = 12 cm (negative)Object distance (u) = 20 cm (negative)Mirror formula:1u+1v=1f\dfrac{1}{\text u} + \dfrac{1}{\text v} = \dfrac{1}{\text f}u1​+v1​=f1​Substituting the values in the formula above we get,−120+1v=−1121v=−112+1201v=−5+3601v=−2601v=−130v=−30 cm- \dfrac{1}{20} + \dfrac{1}{\text v} = - \dfrac{1}{12} \[0.5em] \dfrac{1}{\text v} = - \dfrac{1}{12} + \dfrac{1}{20} \[0.5em] \dfrac{1}{\text v} = \dfrac{ - 5 + 3 }{60} \[0.5em] \dfrac{1}{\text v} = - \dfrac{ 2}{60} \[0.5em] \dfrac{1}{\text v} = - \dfrac{1}{30} \[0.5em] \text v = - 30 \text{ cm} \[0.5em]−201​+v1​=−121​v1​=−121​+201​v1​=60−5+3​v1​=−602​v1​=−301​v=−30 cmHence, the image is formed at a distance of 30 cm in front of the mirror.Magnification (m)=Length of image (I)Length of object (O)=Distance of image (v)Distance of object (u)⇒I2=−3020⇒I=−30×220⇒I=−3 cm\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \[1em] \Rightarrow \dfrac{\text I}{2} = - \dfrac{30}{20} \[0.5em] \Rightarrow \text I = -\dfrac{30 \times 2}{20} \[0.5em] \Rightarrow \text I = - 3 \text { cm}\[0.5em]Magnification (m)=Length of object (O)Length of image (I)​=Distance of object (u)Distance of image (v)​⇒2I​=−2030​⇒I=−2030×2​⇒I=−3 cmHence, length of image = 3 cmImage will be real, inverted and magnified.

Question 9

Question 4An object is placed at 4 cm distance in front of a concave mirror of radius of curvature 24 cm. Find the position of image. Is the image magnified?

Accepted Answer

Given,Radius of curvature (R) = 24 cm (negative)Object distance (u) = 4 cm (negative)Focal length = 12\dfrac{1}{2}21​ x Radius of curvatureSubstituting the values in the formula above, we get,Focal length=12×(−24)Focal length=−12 cm\text{Focal length} = \dfrac{1}{2} \times (- 24) \[0.5em] \text{Focal length} = - 12 \text{ cm}Focal length=21​×(−24)Focal length=−12 cmHence, focal length of the concave mirror = -12 cm.Mirror formula:1u+1v=1f\dfrac{1}{\text u} + \dfrac{1}{\text v} = \dfrac{1}{\text f}u1​+v1​=f1​Now, substituting the values in the mirror formula , we get,−14+1v=−1121v=−112+141v=−1+3121v=2121v=16v=6 cm- \dfrac{1}{4} + \dfrac{1}{\text v} = - \dfrac{1}{12} \[0.5em] \dfrac{1}{\text v} = - \dfrac{1}{12} + \dfrac{1}{4} \[0.5em] \dfrac{1}{\text v} = \dfrac{ - 1 + 3 }{12} \[0.5em] \dfrac{1}{\text v} = \dfrac{2}{12} \[0.5em] \dfrac{1}{\text v} = \dfrac{1}{6} \[0.5em] \text v = 6 \text{ cm} \[0.5em]−41​+v1​=−121​v1​=−121​+41​v1​=12−1+3​v1​=122​v1​=61​v=6 cmThe image is formed 6 cm behind the mirror.Computing linear magnification:Magnification (m)=Length of image (I)Length of object (O)=Distance of image (v)Distance of object (u)⇒m=−6−4⇒m=64⇒m=1.5\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \[1em] \Rightarrow \text m = - \dfrac{6}{- 4} \[0.5em] \Rightarrow \text m = \dfrac{6}{4} \[0.5em] \Rightarrow \text m = 1.5 \[0.5em]Magnification (m)=Length of object (O)Length of image (I)​=Distance of object (u)Distance of image (v)​⇒m=−−46​⇒m=46​⇒m=1.5As, the length of image is 1.5 times the image of object, hence, the image is magnified.

Question 10

Question 6An object 5 cm high is placed at a distance 60 cm in front of a concave mirror of focal length 10 cm. Find (i) the position and (ii) size, of the image.

Accepted Answer

Given,Object height (O) = 5 cmfocal length (f) = 10 cm (negative)Object distance (u) = 60 cm (negative)Mirror formula:1u+1v=1f\dfrac{1}{\text u} + \dfrac{1}{\text v} = \dfrac{1}{\text f}u1​+v1​=f1​Substituting the values in the formula above, we get,−160+1v=−1101v=−110+1601v=−6+1601v=−5601v=−112v=−12 cm- \dfrac{1}{60} + \dfrac{1}{\text v} = - \dfrac{1}{10} \[0.5em] \dfrac{1}{\text v} = - \dfrac{1}{10} + \dfrac{1}{60} \[0.5em] \dfrac{1}{\text v} = \dfrac{ - 6 + 1}{60} \[0.5em] \dfrac{1}{\text v} = - \dfrac{ 5}{60} \[0.5em] \dfrac{1}{\text v} = - \dfrac{1}{12} \[0.5em] \text v = - 12 \text{ cm} \[0.5em]−601​+v1​=−101​v1​=−101​+601​v1​=60−6+1​v1​=−605​v1​=−121​v=−12 cmThe image distance (v) = 12 cm infront of the mirror.Magnification (m)=Length of image (I)Length of object (O)=Distance of image (v)Distance of object (u)I5=−1260I=−12×560⇒I=−1 cm\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \[1em] \dfrac{\text I}{5} = - \dfrac{12}{60} \[0.5em] \text I = -\dfrac{12 \times 5}{60} \[0.5em] \Rightarrow \text I = - 1 \text { cm}\[0.5em]Magnification (m)=Length of object (O)Length of image (I)​=Distance of object (u)Distance of image (v)​5I​=−6012​I=−6012×5​⇒I=−1 cmHence, length of image = 1 cmNegative sign shows that the image will be inverted.

Question 11

Question 7A point light source is kept in front of a convex mirror at a distance of 40 cm. The focal length of the mirror is 40 cm. Find the position of image.

Accepted Answer

Given,u = 40 cm (negative)f = 40 cm (positive)Mirror formula:1u+1v=1f\dfrac{1}{\text u} + \dfrac{1}{\text v} = \dfrac{1}{\text f}u1​+v1​=f1​Substituting the values in the formula above, we get,−140+1v=1401v=140+1401v=1+1401v=2401v=120⇒v=20 cm- \dfrac{1}{40} + \dfrac{1}{\text v} = \dfrac{1}{40} \[0.5em] \dfrac{1}{\text v} = \dfrac{1}{40} + \dfrac{1}{40} \[0.5em] \dfrac{1}{\text v} = \dfrac{1+1}{40} \[0.5em] \dfrac{1}{\text v} = \dfrac{2}{40} \[0.5em] \dfrac{1}{\text v} = \dfrac{1}{20} \[0.5em] \Rightarrow \text v = 20 \text{ cm} \[0.5em]−401​+v1​=401​v1​=401​+401​v1​=401+1​v1​=402​v1​=201​⇒v=20 cmHence, the image is formed 20 cm behind the mirror.

Question 12

Question 8When an object of height 1 cm is kept at a distance 4 cm from a concave mirror, it's erect image of height 1.5 cm is formed at a distance 6 cm behind the mirror. Find the focal length of the mirror.

Accepted Answer

Given,u = 4 cm (negative)v = 6 cm (positive)Mirror formula:1u+1v=1f\dfrac{1}{\text u} + \dfrac{1}{\text v} = \dfrac{1}{\text f}u1​+v1​=f1​Substituting the values in the formula above, we get,−14+16=1f1f=−3+2121f=−112⇒f=−12 cm- \dfrac{1}{4} + \dfrac{1}{6} = \dfrac{1}{\text f} \[0.5em] \dfrac{1}{\text f} = \dfrac{-3+2}{12} \[0.5em] \dfrac{1}{\text f} = - \dfrac{1}{12} \[0.5em] \Rightarrow \text f = - 12 \text{ cm} \[0.5em]−41​+61​=f1​f1​=12−3+2​f1​=−121​⇒f=−12 cmHence, the focal length of concave mirror = 12 cm

Question 13

Question 9An object of length 4 cm is placed in front of a concave mirror at a distance 30 cm. The focal length of mirror is 15 cm. (a) Where will the image form? (b) What will be the length of image?

Accepted Answer

(a) Given,u = 30 cm (negative)f = 15 cm (negative)Mirror formula:1u+1v=1f\dfrac{1}{\text u} + \dfrac{1}{\text v} = \dfrac{1}{\text f}u1​+v1​=f1​Substituting the values in the formula above, we get,−130+1v=−1151v=−115+1301v=−2+1301v=−130⇒v=30 cm- \dfrac{1}{30} + \dfrac{1}{\text v} = - \dfrac{1}{15} \[0.5em] \dfrac{1}{\text v} = - \dfrac{1}{15} + \dfrac{1}{30} \[0.5em] \dfrac{1}{\text v} = \dfrac{ - 2 + 1 }{30} \[0.5em] \dfrac{1}{\text v} = -\dfrac{1}{30} \[0.5em] \Rightarrow \text v = 30 \text{ cm} \[0.5em]−301​+v1​=−151​v1​=−151​+301​v1​=30−2+1​v1​=−301​⇒v=30 cmHence, the image is formed at 30 cm in front of the mirror(b) MagnificationMagnification (m)=Length of image (I)Length of object (O)=Distance of image (v)Distance of object (u)I4=−3030I4=−1⇒I=−4 cm\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \[1em] \dfrac{\text I}{4} = - \dfrac{30}{30} \[0.5em] \dfrac{\text I}{4} = - 1 \[0.5em] \Rightarrow \text I = - 4 \text { cm}\[0.5em]Magnification (m)=Length of object (O)Length of image (I)​=Distance of object (u)Distance of image (v)​4I​=−3030​4I​=−1⇒I=−4 cmHence, length of image = 4 cmNegative sign represents that the image is inverted.

Question 14

Question 10A concave mirror forms a real image of an object placed in front of it at a distance 30 cm, of size three times the size of object. Find (a) the focal length of mirror (b) position of image.

Accepted Answer

Given,(a) Distance of the object (u) = 30 cm (negative)Image height = 3 times the height of objectSo, magnification (m) = 3 (negative for the real image)Magnification (m)=Length of image (I)Length of object (O)=Distance of image (v)Distance of object (u)−3=−v−30⇒v=3×(−30)⇒v=−90 cm\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \[1em] - 3 = - \dfrac{\text v}{- 30} \[0.5em] \Rightarrow \text v = 3 \times (- 30) \[0.5em] \Rightarrow \text v = - 90 \text{ cm} \[0.5em]Magnification (m)=Length of object (O)Length of image (I)​=Distance of object (u)Distance of image (v)​−3=−−30v​⇒v=3×(−30)⇒v=−90 cmHence, the image is formed 90 cm in front of the mirror.Mirror formula:1u+1v=1f\dfrac{1}{\text u} + \dfrac{1}{\text v} = \dfrac{1}{\text f}u1​+v1​=f1​Substituting the values in the mirror formula we get,−130−190=1f1f=−3−1901f=−490⇒f=−22.5 cm- \dfrac{1}{30} - \dfrac{1}{90} = \dfrac{1}{\text f} \[0.5em] \dfrac{1}{\text f} = \dfrac{ - 3 - 1 }{90} \[0.5em] \dfrac{1}{\text f} = -\dfrac{4}{90} \[0.5em] \Rightarrow \text f = -22.5 \text{ cm} \[0.5em]−301​−901​=f1​f1​=90−3−1​f1​=−904​⇒f=−22.5 cmHence, focal length of the mirror = 22.5 cm(b) The image is formed 90 cm in front of the mirror

Question 15

Question 11A concave mirror forms a virtual image of size twice that of the object placed at a distance 5 cm from it. Find (a) the focal length of the mirror (b) position of the image.

Accepted Answer

(a) Given,Distance of the object (u) = 5 cm (negative)Magnification (m) = 2 (positive for the virtual image)Magnification (m)=Length of image (I)Length of object (O)=Distance of image (v)Distance of object (u)2=−v−5⇒v=2×5⇒v=10 cm\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \[1em] 2 = - \dfrac{\text v}{- 5} \[0.5em] \Rightarrow \text v = 2 \times 5 \[0.5em] \Rightarrow \text v = 10 \text{ cm} \[0.5em]Magnification (m)=Length of object (O)Length of image (I)​=Distance of object (u)Distance of image (v)​2=−−5v​⇒v=2×5⇒v=10 cmHence, the image is formed 10 cm behind the mirror.Mirror formula:1u+1v=1f\dfrac{1}{\text u} + \dfrac{1}{\text v} = \dfrac{1}{\text f}u1​+v1​=f1​Substituting the values in the mirror formula, we get,−15+110=−1f−1f=−2+110−1f=−110⇒f=10 cm- \dfrac{1}{5} + \dfrac{1}{10} = - \dfrac{1}{\text f} \[0.5em] - \dfrac{1}{\text f} = \dfrac{ - 2 + 1 }{10} \[0.5em] - \dfrac{1}{\text f} = -\dfrac{1}{10} \[0.5em] \Rightarrow \text f = 10 \text{ cm} \[0.5em]−51​+101​=−f1​−f1​=10−2+1​−f1​=−101​⇒f=10 cmHence, focal length of the mirror = 10 cm(b) The position of the image is 10 cm behind the mirror

Question 16

Question 12The image formed by a convex mirror is of size one-third the size of object. How are u and v related?

Accepted Answer

Magnification (m) = - image distance (v)object distance (u)\dfrac{	ext {image distance (v)}}{	ext{object distance (u)}}object distance (u)image distance (v)​ = - 13\dfrac{1}{3}31​Hence, we get,v = - 13\dfrac{1}{3}31​uOru = -3v

Question 17

Question 13The erect image formed by a concave mirror is of size double the size of object. How are u and v related?

Accepted Answer

Magnification (m) = - image distance (v)object distance (u)\dfrac{	ext {image distance (v)}}{	ext{object distance (u)}}object distance (u)image distance (v)​Given,2 = - vu\dfrac{	ext v}{	ext u}uv​Hence, v = -2u

Question 18

Question 14The magnification for a mirror is -3. How are u and v related?

Accepted Answer

Magnification (m) = - image distance (v)object distance (u)\dfrac{\text {image distance (v)}}{\text{object distance (u)}}object distance (u)image distance (v)​Given,−3=−vu⇒v=3u-3 = -\dfrac{\text{v}}{\text{u}} \[0.5em] \Rightarrow \text{v} = \text{3u}−3=−uv​⇒v=3uHence, v = 3u

Question 19

Question 15How is magnification (m) related to the distance of the object (u) and the distance of the image (v) ?

Accepted Answer

Magnification (m)=Length of image (I)Length of object (O)=Distance of image (v)Distance of object (u)ORm=IO=−vu\text{Magnification (m)} = \dfrac{\text{Length of image (I)}}{\text{Length of object (O)}} \[0.5em] = \dfrac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \[1em] \text{OR} \[1em] m = \dfrac{\text I}{\text O} = - \dfrac{\text v}{\text u}Magnification (m)=Length of object (O)Length of image (I)​=Distance of object (u)Distance of image (v)​ORm=OI​=−uv​where'I' is the length of the image,'O' is the length of the object,'v' is the distance of the image,'u' is the distance of the object.

Reflection of Light

Exercise 7A Very Short Answer Type

Exercise 7C Multiple Choice Type

Exercise 7C Numericals

Exercise 7C Very Short Answer Type