Geometrical Optics — Page 5 | NEET Physics

Q41

A beam of light consisting of red, green and blue colours is incifent on a right angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively. The prism will

A not separate the three colours at all

B separate the red colour part from the green and blue colours

C separate the blue colour part from the red and green colours

D separate all the three colours from one another

Correct Answer

Option B

Solution

for no emergence i $>$ C. sin i $>$ sin C $\Rightarrow$

{1 \over {\sqrt 2 }} > {1 \over \mu }

$\Rightarrow$

\mu > \sqrt 2

= 1.414 As $\mu$ red (= 1.39) < $\mu$ (= 1.414) while $\mu$ green ( = 1.44) and $\mu$ blue (= 1.47) > $\mu$ (= 1.414), so only red colour will be transmitted through face AC while green and blue colours will suffer total internal reflection.

So the prism will separate red colour from the green and blue colours.

Q42

The refracting angle of a prism is A, and refractive index of the material of the prism is cot (A/2). The angle of minimum deviation is

A 90 o

-

A

B 180 o + 2A

C 180 o

-

3A

D 180 o

-

2A

Correct Answer

Option D

Solution

As $\mu$ =

{{\sin \left( {{{A + \delta } \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}

$\Rightarrow$

\cot {A \over 2}

=

{{\sin \left( {{{A + \delta } \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}

$\Rightarrow$

{{\cos {A \over 2}} \over {\sin {A \over 2}}} = {{\sin \left( {{{A + \delta } \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}

$\Rightarrow$

{sin\left( {{\pi \over 2} - {A \over 2}} \right) = \sin \left( {{{A + \delta } \over 2}} \right)}

$\Rightarrow$

{{\pi \over 2} - {A \over 2}}

=

{{A \over 2} + {\delta \over 2}}

$\Rightarrow$ $\delta$ = 180 o - 2A

Q43

Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is

A

-

50 cm

B 50 cm

C

-

20 cm

D

-

25 cm

Correct Answer

Option A

Solution

We know,

{1 \over f} = (\mu - 1)\left( {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right)

For the lens on the left,

{1 \over {{f_1}}} = (1.5 - 1)\left( {{1 \over \infty } - {1 \over { - 20}}} \right) = {1 \over {40}}

For the lensh on the right,

{1 \over {{f_2}}} = (1.5 - 1)\left( {{1 \over { + 20}} - {1 \over \infty }} \right) = {1 \over {40}}

For the oil in the intervening space,

{1 \over {{f_3}}} = (1.7 - 1)\left( {{1 \over { - 20}} - {1 \over {20}}} \right) = - {7 \over {100}}

Hence, if the focal length of the combination is F, then

{1 \over F} = {1 \over {{f_1}}} + {1 \over {{f_2}}} + {1 \over {{f_3}}} = {1 \over {40}} + {1 \over {40}} - {7 \over {100}} = - {1 \over {50}}

i.e., F = $-$ 50 cm

Q44

The angle of a prism is A. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2 A on the first surface returns back through the same path after suffering reflection at the silvered surface. The refractive index

\mu

, of the prism is

A 2sinA

B 2cosA

C

{1 \over 2}

cosA

D tanA

Correct Answer

Option B

Solution

Applying Snell’s law at surface PQ, 1sini = $\mu$ sinr $\mu$ =

{{\sin i} \over {\sin r}}

=

{{\sin 2A} \over {{\mathop{\rm sinA}\nolimits} }}

= 2cosA

Q45

If the focal length of objective lens is increased then magnifying power of

A microscope will increase but that of telescope decrease.

B microscope and telescope both will increase.

C microscope and telescope both will decrease.

D microscope will decrease but that of telescope will increase.

Correct Answer

Option D

Solution

Magnifying power of microscope m =

\left( {{L \over {{f_0}}}} \right)\left( {{D \over {{f_e}}}} \right)

So if f o increases, then m will decrease. Magnifying power of telescope, m =

\left( {{{{f_0}} \over {{f_e}}}} \right)

So if f o increases, then m will increase.

Q46

Two plane mirrors are inclined at 70 o . A ray incident on one mirror at angle,

\theta

after reflection falls on second mirror and is reflected from there parallel to first mirror. The value of

\theta

is

A 45 o

B 30 o

C 55 o

D 50 o

Correct Answer

Option D

Solution

$\theta$ + 40° = 90° $\therefore$ $\theta$ = 90° – 40° = 50°

Q47

A plano convex lens fits exactly into a plano concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices

\mu

1 and

\mu

2 and R is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is

A

{R \over {\left( {{\mu _1} - {\mu _2}} \right)}}

B

{{2R} \over {\left( {{\mu _2} - {\mu _1}} \right)}}

C

{R \over {2\left( {{\mu _1} + {\mu _2}} \right)}}

D

{R \over {2\left( {{\mu _1} - {\mu _2}} \right)}}

Correct Answer

Option A

Solution

Equivalent focal length is

{1 \over {{f_{eq}}}} = {1 \over {{f_1}}} + {1 \over {{f_2}}}

$\Rightarrow$

{1 \over {{f_{eq}}}}

=

\left( {{\mu _1} - 1} \right)\left( {{1 \over \infty } - {1 \over { - R}}} \right)

+

\left( {{\mu _2} - 1} \right)\left( {{1 \over { - R}} - {1 \over \infty }} \right)

$\Rightarrow$ f eq =

{R \over {\left( {{\mu _1} - {\mu _2}} \right)}}

Q48

For a normal eye, the cornea of eye provides a converging power of 40 D and the least converging power of the eye lens behinf the cornea is 20 D. Using this information, the distance between the retina and the cornea-eye lens can be estimated to be

A 1.67 cm

B 1.5 cm

C 5 cm

D 2.5 cm

Correct Answer

Option A

Solution

P cornea = + 40 D P e = + 20 D Total power of combination = 40 + 20 = 60 D Focal length of combination =

{1 \over {60}} \times 100

cm =

{5 \over 3}

cm For minimum converging state of eye lens, u = - $\infty$ v = ? f =

{5 \over 3}

From lens formula,

{1 \over f} = {1 \over v} - {1 \over u}

$\Rightarrow$ v =

{5 \over 3}

cm Distance between retina and cornea-eye lens =

{5 \over 3}

cm = 1.67 cm

Q49

For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index.

A lies between

\sqrt 2

and 1

B lies between 2 and

\sqrt 2

C is less than 1

D is greater than 2

Correct Answer

Option B

Solution

$\mu$ =

{{\sin \left( {{{{\delta _m} + A} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}

=

{{\sin \left( {{{A + A} \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}

=

{{\sin A} \over {\sin \left( {{A \over 2}} \right)}}

=

{{2\sin \left( {{A \over 2}} \right)\cos \left( {{A \over 2}} \right)} \over {\sin \left( {{A \over 2}} \right)}}

=

2\cos \left( {{A \over 2}} \right)

The angle of minimum deviation is given as $\delta$ min = i + e – A for minimum deviation $\delta$ min = A and i = e then 2A = i + i $\Rightarrow$ i = A i min = 0 = A min $\Rightarrow$ $\mu$ min = 2 i max = 90 0 = A max $\Rightarrow$ $\mu$ max =

\sqrt 2

$\therefore$ $\mu$ lies between 2 and

\sqrt 2

.

Q50

A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its enf closer to the pole is 20 cm away from the mirror. The length of the image is

A 10 cm

B 15 cm

C 2.5 cm

D 5 cm

Correct Answer

Option D

Solution

Image position of end A

{1 \over {{v_A}}} + {1 \over { - 20}} = {1 \over { - 10}}

$\Rightarrow$ v A = - 20 cm Image position of the end B

{1 \over {{v_B}}} + {1 \over { - 30}} = {1 \over { - 10}}

$\Rightarrow$ v B = -15 cm Length of the image is L A'B' = |v A | - |v B | = 20 - 15 = 5 cm