Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

Diameter of a plano-convex lens is $$6$$ $$cm$$ and thickness at the center is $$3mm$$. If speed of light in material of lens is $$2 \times {10^8}\,m/s,$$ the focal length of the lens is

A

$$15$$ $$cm$$

B

$$20$$ $$cm$$

C

$$30$$ $$cm$$

D

$$10$$ $$cm$$

$$\therefore$$ $$n = {{Velocity\,\,of\,\,light\,\,in\,\,vacuum} \over {Velocity\,\,of\,\,light\,\,in\,\,medium}}$$

$$\therefore$$ $$n = {3 \over 2}$$

$${3^2} + {\left( {R - 3mm} \right)^2} = {R^2}$$

$$ \Rightarrow {3^2} + {R^2} - 2R\left( {3mm} \right) + {\left( {3mm} \right)^2} = {R^2}$$

$$ \Rightarrow R \approx 15\,cm$$

$${1 \over f} = \left( {{3 \over 2} - 1} \right)\left( {{1 \over {15}}} \right) \Rightarrow f = 30\,cm$$

2

MCQ (Single Correct Answer)

An object $$2.4$$ $$m$$ in front of a lens forms a sharp image on a film $$12$$ $$cm$$ behind the lens. A glass plate $$1$$ $$cm$$ thick, of refractive index $$1.50$$ is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of film?

A

$$7.2$$ $$m$$

B

$$24$$ $$m$$

C

$$3.2$$ $$m$$

D

$$5.6$$ $$m$$

The focal length of the lens

$${1 \over f} = {1 \over \upsilon } - {1 \over u} = {1 \over {12}} + {1 \over {240}}$$

$$ = {{20 + 1} \over {240}} = {{21} \over {240}}$$

$$f = {{240} \over {21}}cm$$

Shift$$ = t\left( {1 - {1 \over \mu }} \right) \Rightarrow 1\left( {1 - {1 \over {3/2}}} \right)$$

$$ = 1 \times {1 \over 3}$$

Now $$v' = 12 - {1 \over 3} = {{35} \over 3}cm$$

Now the object distancce $$u.$$

$${1 \over u} = {3 \over {35}} - {{21} \over {240}} = {1 \over 5}\left[ {{3 \over 7} - {{21} \over {48}}} \right]$$

$${1 \over u} = {1 \over 5}\left[ {{{48 - 49} \over {7 \times 16}}} \right]$$

$$u = - 7 \times 16 \times 5 = - 560cm = - 5.6\,m$$

$${1 \over f} = {1 \over \upsilon } - {1 \over u} = {1 \over {12}} + {1 \over {240}}$$

$$ = {{20 + 1} \over {240}} = {{21} \over {240}}$$

$$f = {{240} \over {21}}cm$$

Shift$$ = t\left( {1 - {1 \over \mu }} \right) \Rightarrow 1\left( {1 - {1 \over {3/2}}} \right)$$

$$ = 1 \times {1 \over 3}$$

Now $$v' = 12 - {1 \over 3} = {{35} \over 3}cm$$

Now the object distancce $$u.$$

$${1 \over u} = {3 \over {35}} - {{21} \over {240}} = {1 \over 5}\left[ {{3 \over 7} - {{21} \over {48}}} \right]$$

$${1 \over u} = {1 \over 5}\left[ {{{48 - 49} \over {7 \times 16}}} \right]$$

$$u = - 7 \times 16 \times 5 = - 560cm = - 5.6\,m$$

3

MCQ (Single Correct Answer)

An electromagnetic wave in vacuum has the electric and magnetic field $$\mathop E\limits^ \to $$and $$\mathop B\limits^ \to $$, which are always perpendicular to each other. The direction of polarization is given by $$\mathop X\limits^ \to $$ and that of wave propagation by $$\mathop k\limits^ \to $$. Then

A

$$\mathop X\limits^ \to ||\mathop B\limits^ \to $$ and $$\mathop X\limits^ \to ||\mathop B\limits^ \to \times \mathop E\limits^ \to $$

B

$$\mathop X\limits^ \to ||\mathop E\limits^ \to $$ and $$\mathop k\limits^ \to ||\mathop E\limits^ \to \times \mathop B\limits^ \to $$

C

$$\mathop X\limits^ \to ||\mathop B\limits^ \to $$ and $$\mathop k\limits^ \to ||\mathop E\limits^ \to \times \mathop B\limits^ \to $$

D

$$\mathop X\limits^ \to ||\mathop E\limits^ \to $$ and $$\mathop k\limits^ \to ||\mathop B\limits^ \to \times \mathop E\limits^ \to $$

as The $$E.M.$$ wave are transverse in nature i.e.,

$$ = {{\overrightarrow k \times \overrightarrow E } \over {\mu \omega }} = \overrightarrow H \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

where $$\overrightarrow H = {{\overrightarrow B } \over \mu }$$

and $${{\overrightarrow k \times \overrightarrow H } \over {\omega \varepsilon }} = - \overrightarrow E \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

$$\overrightarrow k $$ is $$ \bot \,\,\overrightarrow H $$ and $$\overrightarrow k $$ is also $$ \bot $$ to $$\overrightarrow E $$

or In other words $$\overrightarrow X ||\overrightarrow E $$ and $$\overrightarrow k ||\overrightarrow E \times \overrightarrow B $$

$$ = {{\overrightarrow k \times \overrightarrow E } \over {\mu \omega }} = \overrightarrow H \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

where $$\overrightarrow H = {{\overrightarrow B } \over \mu }$$

and $${{\overrightarrow k \times \overrightarrow H } \over {\omega \varepsilon }} = - \overrightarrow E \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

$$\overrightarrow k $$ is $$ \bot \,\,\overrightarrow H $$ and $$\overrightarrow k $$ is also $$ \bot $$ to $$\overrightarrow E $$

or In other words $$\overrightarrow X ||\overrightarrow E $$ and $$\overrightarrow k ||\overrightarrow E \times \overrightarrow B $$

4

MCQ (Single Correct Answer)

In Young's double slit experiment , one of the slit is wider than other, so that amplitude of the light from one slit is double of that from other slit. If $${{\rm I}_m}$$ be the maximum intensity, the resultant intensity $${\rm I}$$ when they interfere at phase difference $$\phi $$ is given by :

A

$${{{I_m}} \over 9}\left( {4 + 5\cos \,\phi } \right)$$

B

$${{{I_m}} \over 3}\left( {1 + 2{{\cos }^2}\,{\phi \over 2}} \right)$$

C

$${{{I_m}} \over 3}\left( {1 + 4{{\cos }^2}\,{\phi \over 2}} \right)$$

D

$${{{I_m}} \over 9}\left( {1 + 8{{\cos }^2}\,{\phi \over 2}} \right)$$

Let $${a_1} = a,\,{I_1} = a_1^2 = {a^2}$$

$${a_2} = 2a,\,{I_2} = a_2^2 = 4{a^2}$$

Therefore $${{\rm I}_2} = 4{{\rm I}_1}$$

$${I_r} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi } $$

$${I_r} = {I_1} + 4{I_1} + 2\sqrt {4I_1^2} \,\cos \phi $$

$$ \Rightarrow {I_r} = 5{I_1} + 4{I_1}\cos \phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Now, $${I_{\max }} = {\left( {{a_1} + {a_2}} \right)^2} = {\left( {a + 2a} \right)^2} = 9{a^2}$$

$${I_{\max }} = 9{I_1} \Rightarrow {I_1} = {{{{\mathop{\rm I}\nolimits} _{max}}} \over 9}$$

Substituting in equation $$\left( 1 \right)$$

$${I_r} = {{5{I_{\max }}} \over 9} + {{4{I_{\max }}} \over 9}\cos \phi $$

$${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 4\cos \phi } \right]$$

$${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 8{{\cos }^2}{\phi \over 2} - 4} \right]$$

$${I_r} = {{{I_{\max }}} \over 9}\left[ {1 + 8{{\cos }^2}{\phi \over 2}} \right]$$

$${a_2} = 2a,\,{I_2} = a_2^2 = 4{a^2}$$

Therefore $${{\rm I}_2} = 4{{\rm I}_1}$$

$${I_r} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi } $$

$${I_r} = {I_1} + 4{I_1} + 2\sqrt {4I_1^2} \,\cos \phi $$

$$ \Rightarrow {I_r} = 5{I_1} + 4{I_1}\cos \phi \,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Now, $${I_{\max }} = {\left( {{a_1} + {a_2}} \right)^2} = {\left( {a + 2a} \right)^2} = 9{a^2}$$

$${I_{\max }} = 9{I_1} \Rightarrow {I_1} = {{{{\mathop{\rm I}\nolimits} _{max}}} \over 9}$$

Substituting in equation $$\left( 1 \right)$$

$${I_r} = {{5{I_{\max }}} \over 9} + {{4{I_{\max }}} \over 9}\cos \phi $$

$${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 4\cos \phi } \right]$$

$${I_r} = {{{I_{\max }}} \over 9}\left[ {5 + 8{{\cos }^2}{\phi \over 2} - 4} \right]$$

$${I_r} = {{{I_{\max }}} \over 9}\left[ {1 + 8{{\cos }^2}{\phi \over 2}} \right]$$

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