Geometrical Optics — Page 4 | NEET Physics

Q31

An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, the displacement of the image will be

A 30 cm away from the mirror

B 36 cm away from the mirror

C 30 cm towards the mirror

D 36 cm towards the mirror

Correct Answer

Option B

Solution

Using mirror formula,

{1 \over f} = {1 \over {{v_1}}} + {1 \over {{u_1}}}

{1 \over { - 15}} = {1 \over {{v_1}}} - {1 \over {40}}

$\Rightarrow$ v 1 = –24 cm When object is displaced by 20 cm towards mirror. Here u 2 = –20 So,

{1 \over f} = {1 \over {{v_2}}} + {1 \over {{u_2}}}

{1 \over { - 15}} = {1 \over {{v_2}}} - {1 \over {20}}

$\Rightarrow$ v 2 = -60 cm So, the image will be shift away from mirror by (60 – 24) cm = 36 cm.

Q32

A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle

\theta

, the spot of the light is found to move through a distance y on the scale. The angle

\theta

is given by

A

{y \over x}

B

{x \over {2y}}

C

{x \over {y}}

D

{y \over {2x}}

Correct Answer

Option D

Solution

As the mirror is rotated θ, the reflected ray rotates 2θ.

Using trigonometry, tan 2θ = y/x θ being small, 2θ = y/x or θ = y/2x

Q33

A thin prism having refracting angle 10 o is made of glass of refractive infex 1.42. This prism is combined with another thin prism of glass of refractive index 1.7. This combination producess dispersion without deviation. The refracting angle of second prism should be

A 6 o

B 8 o

C 10 o

D 4 o

Correct Answer

Option A

Solution

The condition for dispersion without deviation is given as |( $\mu$ – 1)|A = |( $\mu$ ' – 1)| A' $\Rightarrow$ (1.42 – 1) × 10 o = (1.7 – 1)A' $\Rightarrow$ 4.2 = 0.7A' or A' = 6 o

Q34

Two identical glass (

\mu

g = 3/2) equiconvex lenses of focal length

f

each are kept in contact. The space between the two lenses is filled with water

\left( {{\mu _w} = 4/3} \right)

. The focal length of the combination is

A

f/3

B

f

C

4f/3

D

3f/4

Correct Answer

Option D

Solution

Focal length (f) of glass convex lens is given by

{1 \over f} = \left( {{\mu _g} - 1} \right)\left( {{2 \over R}} \right)

$\Rightarrow$

{1 \over f} = \left( {{3 \over 2} - 1} \right)\left( {{2 \over R}} \right)

$\Rightarrow$ f = R Focal length (f 1 ) of water filled concave lens is given by

{1 \over {{f_1}}} = \left( {{\mu _w} - 1} \right)\left( { - {2 \over R}} \right)

$\Rightarrow$

{1 \over {{f_1}}} = \left( {{4 \over 3} - 1} \right)\left( { - {2 \over R}} \right)

=

{ - {2 \over {3R}}}

=

{ - {2 \over {3f}}}

Equivalent focal length (f eq ) of lens system

{1 \over {{f_{eq}}}} = {1 \over f} - {2 \over {3f}} + {1 \over f}

=

{4 \over {3f}}

$\therefore$ f eq =

{{3f} \over 4}

Q35

A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes . In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use. will be

A convex, + 2.25 diopter

B concave,

-

0.25 diopter

C concave,

-

0.2 diopter

D convex, + 0.15 diopter

Correct Answer

Option B

Solution

Given u = 400 cm = 4 m v = $\infty$ and f = ? Since,

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

=

{1 \over \infty } - {1 \over 4}

so, f = – 4 m Hence, lens will be concave. Now P =

{1 \over f}

=

{1 \over { - 4}}

= - 0.25 D

Q36

An air bubble in a glass slab with refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is

A 8

B 10

C 12

D 16

Correct Answer

Option C

Solution

Seeing from one end, h 1 = $\mu$ × (h – b) = 3/2 × 5 = 15/2 cm From other end of the slab, h 2 = $\mu$ × h = 3/2 × 3 = 9/2 cm Now total height, (15/2 + 9/2) = 24/2 = 12 cm

Q37

A astronomical telescope has objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance

A 50.0 cm

B 54.0 cm

C 37.3 cm

D 46.0 cm

Correct Answer

Option B

Solution

Given: Focal length of objective, f 0 = 40cm Focal length of eye–piece f e = 4 cm image distance, u 0 = 200 cm Using lens formula for objective lens

{1 \over {{v_0}}} - {1 \over {{u_0}}} = {1 \over {{f_0}}}

{1 \over {{v_0}}} - {1 \over { - 200}} = {1 \over {40}}

$\Rightarrow$

{1 \over {{v_0}}} = {1 \over {40}} - {1 \over {200}}

=

{4 \over {200}}

$\Rightarrow$ v 0 = 50 cm Tube length(

l

) = Distance between lenses = v o + f e = 50 + 4 = 54 cm

Q38

Match the corresponding entries of column 1 with column 2. [Where m is the magnification produced by the mirror] <br><br> <table class=tg> <tbody><tr> <th class=tg-5qg7 colspan=2>Column 1</th> <th class=tg-5qg7></th> <th class=tg-hjji colspan=2>Column 2</th> </tr> <tr> <td class=tg-5qg7>(A)</td> <td class=tg-dzkb>m = - 2</td> <td class=tg-5qg7></td> <td class=tg-hjji>(p)</td> <td class=tg-84g4>Convex mirror</td> </tr> <tr> <td class=tg-5qg7>(B)</td> <td class=tg-dzkb>m = -

{1 \over 2}

</td> <td class=tg-5qg7></td> <td class=tg-hjji>(q)</td> <td class=tg-84g4>Concave mirror</td> </tr> <tr> <td class=tg-5qg7>(C)</td> <td class=tg-dzkb>m = +2</td> <td class=tg-5qg7></td> <td class=tg-hjji>(r)</td> <td class=tg-84g4>Real image</td> </tr> <tr> <td class=tg-hjji>(D)</td> <td class=tg-84g4>m = +

{1 \over 2}

</td> <td class=tg-hjji></td> <td class=tg-hjji>(s)</td> <td class=tg-84g4>Virtual image</td> </tr> </tbody></table>

A A

\to

p and s; B

\to

q and r; C

\to

q and s; D

\to

q and r

B A

\to

r and s; B

\to

q and s; C

\to

q and r; D

\to

p and s

C A

\to

q and r; B

\to

q and r; C

\to

q and s; D

\to

p and s

D A

\to

p and r; B

\to

p and s; C

\to

p and q; D

\to

r and s

Correct Answer

Option C

Solution

Magnification in the mirror, m =

- {v \over u}

m = –2 $\Rightarrow$ v = 2u As v and u have same signs so the mirror is concave and image formed is real.

m = - {1 \over 2}

$\Rightarrow$ v =

{u \over 2}

$\Rightarrow$ Concave mirror and real image. m = + 2 $\Rightarrow$ v = –2u As v and u have different signs but magnification is 2 so the mirror is concave and image formed is virtual.

m = + {1 \over 2}

$\Rightarrow$ v =

- {u \over 2}

As v and u have different signs with magnification

{1 \over 2}

so the mirror is convex and image formed is virtual.

Q39

The angle of incidence for a ray of light at a refracting surface of a prism is 45 o . The angle of prism is 60 o . If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are

A

{45^o};\sqrt 2

B

{30^o};{1 \over {\sqrt 2 }}

C

{45^o};{1 \over {\sqrt 2 }}

D

{30^o};\sqrt 2

Correct Answer

Option D

Solution

i =

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

$\Rightarrow$ 45 o =

{{{\delta _m} + 60^\circ } \over 2}

$\Rightarrow$

{{\delta _m} = 30^\circ }

sin45º = $\mu$ sin30º $\Rightarrow$

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

$\Rightarrow$ $\mu$ =

{\sqrt 2 }

Q40

In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is

I

. The magnification of the telescope is

A

{{L + I} \over {L - I}}

B

{L \over I}

C

{L \over I} + 1

D

{L \over I} - 1

Correct Answer

Option B

Solution

Magnification by eye piece m =

{f \over {f + u}}

$\Rightarrow$

- {I \over L} = {{{f_e}} \over {{f_e} + \left[ { - \left( {{f_0} + {f_e}} \right)} \right]}}

= -

{{{f_e}} \over {{f_0}}}

$\Rightarrow$

{I \over L} = {{{f_e}} \over {{f_0}}}

$\therefore$ Magnification, M =

{{{f_0}} \over {{f_e}}}

=

{L \over I}