Q: An object of size 7 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 focused image can be obtained ? Find the size and the nature of the image.

Mirror formula : \$\dfrac{1}{v} +\dfrac{1}{u}=\dfrac{1}{f}\$...........(1)So, \$u=-27cm\$Concave mirror have negative focal length.\$f=-18cm\$Putting the values in (1),\$\dfrac{1}{v}+\dfrac{1}{-27}=\dfrac{1}{-18}\$We get , \$v=-54cm\$Also,Magnification = \$\dfrac{h_2}{h_1}=\dfrac{-v}{u}\$\$\dfrac{h_2}{7.0}=-\dfrac{-54}{-27}\$\$h_2=-14.0 cm\$Image will be real and inverted and will be of size 14 cm.

Q: The absolute refractive index of glass and water are \$3/2\$ and \$4/3\$, respectively. If the speed of light in glass is \$2 \times 10^8\$ m/s, calculate the speed of light in :(a) vacuum(b) water

Given : \$n_{water} = \dfrac{4}{3}\$ \$n_{glass} = \dfrac{3}{2}\$ \$v_{glass} = 2\times 10^8\$ m/s \$n_{air} = 1\$Speed of light in a medium \$v_{medium} = \dfrac{c}{n}\$ where \$c\$ and \$n\$ are the speed of light in vacuum and refractive index of medium respectively.(a) : \$v_{glass} = \dfrac{c}{n_{glass}}\$\$\therefore\$ \$2\times 10^8 = \dfrac{c}{(3/2)}\$ \$\implies c = 3\times 10^8\$ m/s(b) : \$v_{water} = \dfrac{c}{n_{water}}\$\$\therefore\$ \$v_{water} = \dfrac{3\times 10^8}{(4/3)} = 2.26\times 10^8\$ m/s

Q: A \$4.5\ cm\$ needle is placed \$12\ cm\$ away from a convex mirror of focal length \$15\ cm\$. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Given: Height of the needle, \$h_1 = 4.5 cm\$.Object distance, \$u = −12 cm\$.The focal length of the convex mirror, \$f=15cm\$.Image distance, \$ v\$ The value of \$v\$ can be obtained using the mirror formula.\$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\$\$\dfrac{1}{v}+\dfrac{1}{-12}=\dfrac{1}{15}\$\$\dfrac{1}{v} = \dfrac{1}{12}+\dfrac{1}{15}\$\$\dfrac{1}{v} = \dfrac{9}{60}\$\$\therefore v \approx 6.7cm\$Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other side of the mirror.The image size is given by the magnification formula.\$m=\dfrac{h^{'}}{h}=-\dfrac{v}{u}\$\$h'=\dfrac{6.7\times4.5}{12}\$\$\Rightarrow h'=+2.5cm\$So, \$m=\dfrac{2.5}{4.5}\$\$m=0.56\$The height of the image is \$2.5 cm\$. The positive sign indicates that the image is erect, virtual, and diminished. If the needle is moved farther from the mirror, the size of the image will reduce gradually.

Q: A concave lens has a focal length of \$20\ cm\$. At what distance from a lens a \$5\ cm\$ tall object should be placed, so that it forms an image at \$15\ cm\$ from the lens? Also, calculate the size of the image formed.

According to the question,Image distance, \$v = -15\ cm\$ (negative due to convex lens)Focal length, \$f = -20\ cm\$Let the object distance be \$ u \$.By lens formula:\$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\[4pt]\$\$\Rightarrow \dfrac{1}{-15\ cm} - \dfrac{1}{u} = \dfrac{1}{-20\ cm} \\[4pt]\$\$\Rightarrow \dfrac{1}{u} = \dfrac{1}{20\ cm} + \dfrac{1}{-15\ cm} \\[4pt]\$\$\Rightarrow \dfrac{1}{u} = \dfrac{3 - 4}{60\ cm} = \dfrac{-1}{60\ cm} \\[4pt]\$\$\Rightarrow u = -60\ cm\$Therefore, object is placed at \$60\ cm\$ away from the lens, on the same side as image.Now,Height of object, \$h_1 = 5cm\$Magnification, \$m = \dfrac{h_2}{h_1} = \dfrac{v}{u}\$Putting value of \$ v \$ and \$ u \$:Magnification, \$m = \dfrac{h_2}{5\ cm} = \dfrac{-15\ cm}{-60\ cm}\$\$\Rightarrow \dfrac{h_2}{5\ cm}=\dfrac{1}{4}\$\$\Rightarrow h_2=\dfrac{5\ cm}{4} = 1.25\ cm\$Thus, the height of the image is \$1.25\ cm\$ and the positive sign means the image is virtual and erect.

Q: A Concave mirror produces three times magnified real image of an object placed at \$10cm\$ infront of it . Where is the image located?

Given magnification \$m=3\$and object distance \$u=10\$ cm and we know,Magnification \$m=\dfrac{-v}{u}\$\$-v=3\times -10\$\$v=30\$ cmThus the image is formed at distance of 30 cm.

Q: A silicon optical fibre with a core diameter large enough has a core refractive index of 1.50 and a cladding refractive index 1.47. Determine(i) the critical angle at the core cladding interface,(ii) the numerical aperture for the fibre(iii) the acceptance angle in air for the fibre.

Here,\${ \mu }_{ 1 }=1.50;{ \mu }_{ 2 }=1.47;\$\${ \mu }_{ 0 }=1\$(i) Critical angle \${ \theta }_{ c }\$ at the core-cladding interface is given by \${ \theta }_{ c }={ sin }^{ -1 }\dfrac { 1.47 }{ 1.50 } =78.{ 5 }^{ 0 }\$(ii) Numerical aperture,\$NA={ \left( { { \mu }_{ 1 } }^{ 2 }-{ { \mu }_{ 2 } }^{ 2 } \right) }^{ \frac { 1 }{ 2 } }\$ \$={ \left[ { \left( 1.50 \right) }^{ 2 }-{ \left( 1.47 \right) }^{ 2 } \right] }^{ \frac { 1 }{ 2 } }={ \left( 2.25-2.16 \right) }^{ 1/2 }=0.30\$(iii) Acceptance angle \${ \theta }_{ a }={ sin }^{ -1 }\left( NA \right) \$\$={ sin }^{ -1 }\left( 0.30 \right) ={ 17.4 }^{ 0 }\$

Q: The image formed by a plane mirror is always

A \$ \bf{plane\ mirror} \$ always forms an image which is \$ \bf{virtual} \$ in nature (meaning that light rays meet behind the mirror or that light rays do not actually come from the image), \$ \bf{erect} \$ and of the \$ \bf{same\ shape\ and\ size} \$ as that of the object, it is reflecting.Hence, the correct answer is OPTION B.

Question 1

An object of size 7 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 focused image can be obtained ? Find the size and the nature of the image.

Accepted Answer

Mirror formula : $\dfrac{1}{v} +\dfrac{1}{u}=\dfrac{1}{f}$...........(1)So, $u=-27cm$Concave mirror have negative focal length.$f=-18cm$Putting the values in (1),$\dfrac{1}{v}+\dfrac{1}{-27}=\dfrac{1}{-18}$We get , $v=-54cm$Also,Magnification = $\dfrac{h_2}{h_1}=\dfrac{-v}{u}$$\dfrac{h_2}{7.0}=-\dfrac{-54}{-27}$$h_2=-14.0 cm$Image will be real and inverted and will be of size 14 cm.

Question 2

The absolute refractive index of glass and water are $3/2$ and $4/3$, respectively. If the speed of light in glass is $2 \times 10^8$ m/s, calculate the speed of light in :(a) vacuum(b) water

Accepted Answer

Given :    $n_{water} = \dfrac{4}{3}$           $n_{glass} = \dfrac{3}{2}$             $v_{glass} = 2\times 10^8$  m/s            $n_{air} = 1$Speed of light in a medium        $v_{medium} = \dfrac{c}{n}$         where  $c$ and $n$ are the speed of light in vacuum and refractive index of medium respectively.(a) :     $v_{glass} = \dfrac{c}{n_{glass}}$$\therefore$     $2\times 10^8  = \dfrac{c}{(3/2)}$                        $\implies c  = 3\times 10^8$  m/s(b) :     $v_{water} = \dfrac{c}{n_{water}}$$\therefore$     $v_{water}  = \dfrac{3\times 10^8}{(4/3)}  = 2.26\times 10^8$  m/s

Question 3

A $4.5\ cm$ needle is placed $12\ cm$ away from a convex mirror of focal length $15\ cm$. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.

Accepted Answer

Given: Height of the needle, $h_1 = 4.5 cm$.Object distance, $u = &#8722;12 cm$.The focal length of the convex mirror, $f=15cm$.Image distance, $ v$ The value of $v$ can be obtained using the mirror formula.$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$$\dfrac{1}{v}+\dfrac{1}{-12}=\dfrac{1}{15}$$\dfrac{1}{v} = \dfrac{1}{12}+\dfrac{1}{15}$$\dfrac{1}{v} = \dfrac{9}{60}$$\therefore v \approx 6.7cm$Hence, the image of the needle is 6.7 cm away from the mirror. Also, it is on the other&#10;side of the mirror.The image size is given by the magnification formula.$m=\dfrac{h^{'}}{h}=-\dfrac{v}{u}$$h'=\dfrac{6.7\times4.5}{12}$$\Rightarrow h'=+2.5cm$So, $m=\dfrac{2.5}{4.5}$$m=0.56$The height of the image is $2.5 cm$. The positive sign indicates that the image is erect,&#10;virtual, and diminished.&#10;If the needle is moved farther from the mirror, the size of the image will reduce gradually.

Question 4

A concave lens has a focal length of $20\ cm$. At what distance from a lens a $5\ cm$  tall object should be placed, so that it forms an image at  $15\ cm$ from the lens? Also, calculate the size of the image formed.

Accepted Answer

According to the question,Image distance, $v = -15\ cm$ (negative due to convex lens)Focal length, $f = -20\ cm$Let the object distance be $ u $.By lens formula:$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \[4pt]$$\Rightarrow \dfrac{1}{-15\ cm} - \dfrac{1}{u} = \dfrac{1}{-20\ cm} \[4pt]$$\Rightarrow \dfrac{1}{u} = \dfrac{1}{20\ cm} + \dfrac{1}{-15\ cm} \[4pt]$$\Rightarrow \dfrac{1}{u} = \dfrac{3 - 4}{60\ cm} = \dfrac{-1}{60\ cm} \[4pt]$$\Rightarrow  u = -60\ cm$Therefore, object is placed at $60\ cm$ away from the lens, on the same side as image.Now,Height of object, $h_1 = 5cm$Magnification, $m = \dfrac{h_2}{h_1} = \dfrac{v}{u}$Putting value of $ v $ and $ u $:Magnification, $m = \dfrac{h_2}{5\ cm} = \dfrac{-15\ cm}{-60\ cm}$$\Rightarrow \dfrac{h_2}{5\ cm}=\dfrac{1}{4}$$\Rightarrow h_2=\dfrac{5\ cm}{4} = 1.25\ cm$Thus, the height of the image is $1.25\ cm$ and the positive sign means the image is virtual and erect.

Question 5

A Concave mirror produces three times magnified real image of an object placed at $10cm$ infront of it . Where is the image located?

Accepted Answer

Given magnification $m=3$and object distance $u=10$ cm and we know,Magnification $m=\dfrac{-v}{u}$$-v=3\times -10$$v=30$ cmThus the image is formed at distance of 30 cm.

Question 6

A silicon optical fibre with a core diameter large enough has a core refractive index of 1.50 and a cladding refractive index 1.47. Determine(i) the critical angle at the core cladding interface,(ii) the numerical aperture for the fibre(iii) the acceptance angle in air for the fibre.

Accepted Answer

Here,${ \mu  }_{ 1 }=1.50;{ \mu  }_{ 2 }=1.47;$${ \mu  }_{ 0 }=1$(i) Critical angle ${ \theta  }_{ c }$ at the core-cladding interface is given by     ${ \theta  }_{ c }={ sin }^{ -1 }\dfrac { 1.47 }{ 1.50 } =78.{ 5 }^{ 0 }$(ii) Numerical aperture,$NA={ \left( { { \mu  }_{ 1 } }^{ 2 }-{ { \mu  }_{ 2 } }^{ 2 } \right)  }^{ \frac { 1 }{ 2 }  }$       $={ \left[ { \left( 1.50 \right)  }^{ 2 }-{ \left( 1.47 \right)  }^{ 2 } \right]  }^{ \frac { 1 }{ 2 }  }={ \left( 2.25-2.16 \right)  }^{ 1/2 }=0.30$(iii) Acceptance angle ${ \theta  }_{ a }={ sin }^{ -1 }\left( NA \right) $$={ sin }^{ -1 }\left( 0.30 \right) ={ 17.4 }^{ 0 }$

Question 7

At noon the sun appears white because:

Accepted Answer

At noon because the sun is overhead, the light is scattered the least and hence appears white. When it is overhead, it has lesser air to travel through and the scattering from dust and other particles is reduced if the distance to be travelled in air is reduced.

Question 8

If an object is placed at a distance of $0.5\,m$ in front of a plane mirror, the distance between the object and the image formed by the mirror will be

Accepted Answer

The image formed by the mirror is at same distance behind the mirror as object is in front of it, nowIf distance of object from mirror = 0.5 mThen distance of image from mirror = 0.5 mSo, distance between object and mirror = 0.5 m+ 0.5m = 1m

Question 9

State the characteristics of the image formed by a plane mirror.

Accepted Answer

(i) Plane mirror forms an erect image.(ii) It forms a virtual image.(iii) Size of the image is same as that of the object.(iv) Image is formed at the same distance behind the mirror as the object stands in front of it.(v)Image formed is a laterally inverted image i.e., right hand side of the object seems to be the left hand side and vice-versa.

Question 10

The image formed by a plane mirror is always

Accepted Answer

A $ \bf{plane\ mirror} $ always forms an image which is $ \bf{virtual} $ in nature (meaning that light rays meet behind the mirror or that light rays do not actually come from the image), $ \bf{erect} $ and of the $ \bf{same\ shape\ and\ size} $ as that of the object, it is reflecting.Hence, the correct answer is OPTION B.