CBSE Questions for Class 10 Physics Light Reflection And Refraction Quiz 11 - MCQExams.com

Since, it is given that angle of incidence is equal to the critical angle

Therefore,

  $$ \sin \,i\,=\,\dfrac{1}{n}\,\,=\,\dfrac{1}{\sqrt{2}} $$

 $$ i\,=\,\,45{}^\circ  $$

 $$  $$According to laws of refraction

  $$ \dfrac{\sin i}{\sin r}\,=\,n\,\,(refractive\,index) $$

 $$ sin\,r\,=\,\dfrac{\sin \,i}{n}\,=\,\,\frac{1}{2} $$

 $$ r\,=\,30{}^\circ  $$

Let $$\theta $$ is the angle between R₃ and R₄

So, $$ \theta \,=\,180{}^\circ \,-\,(i\,+\,r)\, $$

 $$ \,\,\,\,\,=\,180\,-\,(\,45\,+\,30\,) $$

 $$ \,\,\,\,\,\,=\,\,105{}^\circ  $$

The magnification is given as,

$$m = \dfrac{{ - v}}{u}$$

$$2 = \dfrac{{ - v}}{u}$$

$$v =  - 2u$$

Ignoring the sign and using mirror formula, we get

$$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$$

$$\dfrac{1}{{2u}} + \dfrac{1}{u} = \dfrac{1}{f}$$

$$\dfrac{{1 + 2}}{{2u}} = \dfrac{1}{f}$$

$$u = \dfrac{{3f}}{2}$$

Here, difference between object distance and image distance is also$$u$$.

Given that,

The object distance $$u=-6\,cm$$

Now, magnification is

  $$ m=\dfrac{I}{O} $$

 $$ m=\dfrac{f}{f-u} $$

 $$ \dfrac{I}{6}=\dfrac{-f}{-f-\left( -4f \right)} $$

 $$ I=-2\,cm $$

Hence, the length of image is -$$2\ cm$$

The size of the image is equal to size of the object for plane mirror.

So, magnification is equal to ratio of image height to object height so it will subsequently be $$1$$ .

The incident ray $$1$$ will be reflected at an equal angle as reflected ray $$2.$$

By laws of reflection, $$i=r=60^o$$ 

The refracted ray $$3$$ is perpendicular to reflected ray $$2.$$

So by geometry, the angle of refraction $$r'=30^o$$

Now by Snell's law,

$$\mu_{air} \ sin \ i= \mu_{glass} \ sin \ r'$$

$$(1) \times sin \ 60^o= \mu_{glass} \times sin \ 30^o$$

$$\dfrac{\sqrt3}{2}=\mu_{glass} \times \dfrac{1}{2}$$

$$\mu_{glass}=\sqrt3$$

Option A is the answer.

Suppose $${\eta _1}$$ and $${\eta _2}$$ is the refractive index of medium $$1$$ and $$2$$.

The refractive index of the medium $$1$$ with respect to the medium $$2$$ is given as

$${\eta _{12}} = \dfrac{{{\eta _1}}}{{{\eta _2}}}$$

Similarly the refractive index of the medium $$2$$ with respect to the medium $$1$$ is given as

$${\eta _{21}} = \dfrac{{{\eta _2}}}{{{\eta _1}}}$$

Hence,

$${\eta _{12}} \times {\eta _{21}} = \dfrac{{{\eta _1}}}{{{\eta _2}}} \times \dfrac{{{\eta _2}}}{{{\eta _1}}}$$

$${\eta _{12}} \times {\eta _{21}} = 1$$.

Given,

The size of the image of object A and B is the same, hence

$${h_A}^\prime  = {h_B}^\prime $$

The size of object A is four times that of B, hence

$${h_A} = 3{h_B}$$

Now the magnification of object A can be given as

$${m_A} = \dfrac{{{h_A}^\prime }}{{{h_A}}}$$                                           …… (1)

Similarly for object B

$${m_B} = {\text{ }}\dfrac{{{h_B}^\prime }}{{{h_B}}}$$                                           ……. (2)

From equation (1) and (2) the ratio of magnification is given as

$$\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{{{h_A}^\prime }}{{{h_A}}}{\text{ }} \times {\text{ }}\dfrac{{{h_B}}}{{{h_B}^\prime }} \Rightarrow \dfrac{{{m_A}}}{{{m_B}}} = \dfrac{{{h_A}^\prime }}{{3{h_B}}}{\text{ }} \times {\text{ }}\dfrac{{{h_B}}}{{{h_A}^\prime }}$$

$$\therefore \dfrac{{{m_A}}}{{{m_B}}} = \dfrac{1}{3}$$                                       ……. (3)

Now the magnification formula is given as

$$m = \dfrac{f}{{f - u}}$$

Therefore equation (3) becomes

$$\dfrac{{{m_A}}}{{{m_B}}} = \dfrac{{f - {u_B}}}{{f - {u_A}}} \Rightarrow \dfrac{1}{3} = \dfrac{{ - 7.5 - {u_B}}}{{ - 7.5 + 30}}$$

$$ - {u_B} = \dfrac{{22.5}}{3} + 10 \Rightarrow {u_B} =  - 17.5cm$$

Hence the object B is placed at $$-17.5\;cm$$ from the mirror.