Joint Entrance Examination

Graduate Aptitude Test in Engineering

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1

MCQ (Single Correct Answer)

The box of a pin hole camera, of length $$L,$$ has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength $$\lambda $$ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say $${b_{\min }}$$) when :

A

$$a = \sqrt {\lambda L} \,$$ and $${b_{\min }} = \sqrt {4\lambda L} $$

B

$$a = {{{\lambda ^2}} \over L}$$ and $${b_{\min }} = \sqrt {4\lambda L} $$

C

$$a = {{{\lambda ^2}} \over L}$$ and $${b_{\min }} = \left( {{{2{\lambda ^2}} \over L}} \right)$$

D

$$a = \sqrt {\lambda 1} $$ and $${b_{\min }} = \left( {{{2{\lambda ^2}} \over L}} \right)$$

Given geometrical spread $$=a$$

Diffraction spread $$ = {\lambda \over a} \times L = {{\lambda L} \over a}$$

The sum $$b = a + {{\lambda L} \over a}$$

For $$b$$ to be minimum $${{db} \over {da}} = 0$$ $${d \over {da}}\left( {a + {{\lambda L} \over a}} \right) = 0$$

$$a = \sqrt {\lambda L} $$

$$b\min = \sqrt {\lambda L} + \sqrt {\lambda L} = 2\sqrt {\lambda L} = \sqrt {4\lambda L} $$

Diffraction spread $$ = {\lambda \over a} \times L = {{\lambda L} \over a}$$

The sum $$b = a + {{\lambda L} \over a}$$

For $$b$$ to be minimum $${{db} \over {da}} = 0$$ $${d \over {da}}\left( {a + {{\lambda L} \over a}} \right) = 0$$

$$a = \sqrt {\lambda L} $$

$$b\min = \sqrt {\lambda L} + \sqrt {\lambda L} = 2\sqrt {\lambda L} = \sqrt {4\lambda L} $$

2

MCQ (Single Correct Answer)

Monochromatic light is incident on a glass prism of angle $$A$$. If the refractive index of the material of the prism is $$\mu $$, a ray, incident at an angle $$\theta $$. on the face $$AB$$ would get transmitted through the face $$AC$$ of the prism provided :

A

$$\theta > {\cos ^{ - 1}}\left[ {\mu \,\sin \left( {A + {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]$$

B

$$\theta < {\cos ^{ - 1}}\left[ {\mu \,\sin \left( {A + {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]$$

C

$$\theta > si{n^{ - 1}}\left[ {\mu \,\sin \left( {A - {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]$$

D

$$\theta < si{n^{ - 1}}\left[ {\mu \,\sin \left( {A - {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]$$

When $${r_2} = C,\,\angle {N_2}Rc = {90^ \circ }$$

Where $$C = $$ critical angle

As $$\sin C = {1 \over v} = \sin {r_2}$$

Applying snell's law at $$'R'$$

$$\mu \sin {r_2} = 1\sin {90^ \circ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Applying snell's law at $$'Q'$$

$$1 \times \sin \theta = \mu \sin {r_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

But $${r_1} = A - {r_2}$$

So, $$\sin \theta = \mu \sin \left( {A - {r_2}} \right)$$

$$\theta = \mu \sin A\cos {r_2} - \cos \,A\,\,\,\,\,...\left( {iii} \right)$$ [using $$(i)$$]

From $$(1)$$

$$\cos \,{r_2} = \sqrt {1 - {{\sin }^2}{r_2}} = \sqrt {1 - {1 \over {{\mu ^2}}}} \,\,\,\,\,\,\,...\left( {iv} \right)$$

By eq.$$(iii)$$ and $$(iv)$$

$$\sin \theta = \mu \sin A\sqrt {1 - {1 \over {{\mu ^2}}}} - \cos A$$

on further solving we can show for ray not to transmitted through face $$AC$$

$$\theta = {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right)} \right.} \right]$$

So, for transmission through face $$AC$$

$$\theta > {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right.} \right)} \right]$$

Where $$C = $$ critical angle

As $$\sin C = {1 \over v} = \sin {r_2}$$

Applying snell's law at $$'R'$$

$$\mu \sin {r_2} = 1\sin {90^ \circ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$

Applying snell's law at $$'Q'$$

$$1 \times \sin \theta = \mu \sin {r_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

But $${r_1} = A - {r_2}$$

So, $$\sin \theta = \mu \sin \left( {A - {r_2}} \right)$$

$$\theta = \mu \sin A\cos {r_2} - \cos \,A\,\,\,\,\,...\left( {iii} \right)$$ [using $$(i)$$]

From $$(1)$$

$$\cos \,{r_2} = \sqrt {1 - {{\sin }^2}{r_2}} = \sqrt {1 - {1 \over {{\mu ^2}}}} \,\,\,\,\,\,\,...\left( {iv} \right)$$

By eq.$$(iii)$$ and $$(iv)$$

$$\sin \theta = \mu \sin A\sqrt {1 - {1 \over {{\mu ^2}}}} - \cos A$$

on further solving we can show for ray not to transmitted through face $$AC$$

$$\theta = {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right)} \right.} \right]$$

So, for transmission through face $$AC$$

$$\theta > {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right.} \right)} \right]$$

3

MCQ (Single Correct Answer)

Assuming human pupil to have a radius of $$0.25$$ $$cm$$ and a comfortable viewing distance of $$25$$ $$cm$$, the minimum separation between two objects that human eye can resolve at $$500$$ $$nm$$ wavelength is :

A

$$100\,\mu m$$

B

$$300\,\mu m$$

C

$$1\,\mu m$$

D

$$30\,\mu m$$

$$\sin \theta = {{0.25} \over {25}} = {1 \over {100}}$$

Resolving power $$ = {{1.22\lambda } \over {2\mu \sin \theta }} = 30\,\mu m.$$

Resolving power $$ = {{1.22\lambda } \over {2\mu \sin \theta }} = 30\,\mu m.$$

4

MCQ (Single Correct Answer)

On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens' principle leads us to conclude that as it travels, the light beam :

A

bends down wards

B

bends upwards

C

becomes narrower

D

goes horizontally without any deflection

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