Geometrical Optics — Page 5 | JEE Physics

Q41

A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15cm from a converging lens of magnitude of focal length 20cm. A beam of parallel light falls on the diverging lens. The final image formed is:

A real and at a distance of 6 cm from the convergent lens.

B real and at a distance of 40 cm from convergent lens.

C virtual and at a distance of 40 cm from convergent lens.

D real and at a distance of 40 cm from the divergent lens.

Correct Answer

Option B

Solution

As parallel beam incident on diverging lens so the image will be formed at the focus of diverging lens. $\therefore$ v = –25 cm The image formed by diverging lens is used as an object for converging lens.

So for converging lens u = –25 – 15 = –40 cm, f = 20 cm So Final image formed by converging lens

{1 \over {20}} = {1 \over V} - {1 \over { - 40}}

$\Rightarrow$ V = 40 cm V is positive so image will be real and will form at right side of converging lens at 40 cm.

Q42

A ray of light is incident at an angle of 60o on one face of a prism of angle 30o. The emergent ray of light makes an angle of 30o with incident ray. The angle made by the emergent ray with second face of prism will be :

A 0o

B 90o

C 45o

D 30o

Correct Answer

Option B

Solution

Given :

{i_1} = 60^\circ

;

A = 30^\circ

Then angle of deviation is given by

\delta = 1 - A = 60^\circ - 30^\circ = 30^\circ

We know that,

{i_1} + {i_2} = A + \delta \Rightarrow 60^\circ + {i_2} = 30^\circ + 30^\circ \Rightarrow {i_2} = 0

It means emergent ray leaves face AC normally.

Q43

A concave mirror for face viewing has focal length of 0.4 m. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is :

A 0.24 m

B 0.32 m

C 1.60 m

D 0.16 m

Correct Answer

Option B

Solution

m = {f \over {f - u}}

5 = {{ - 40} \over { - 40 - u}};\,u = - 32cm

Q44

A convex lens (of focal length 20 cm) and a concave mirror, having their principal axes along the same lines, are kept 80 cm apart from each other. The concave mirror is to the right of the convex lens. When an object is kept at a distance of 30 cm to the left of the convex lens, its image remains at the same position even if the concave mirror is removed. The maximum distance of the object for which this concave mirror, by itself would produce a virtual image would be :-

A 25 cm

B 10 cm

C 20 cm

D 30 cm

Correct Answer

Option B

Solution

Image formed by lens

{1 \over v} - {1 \over u} = {1 \over f};{1 \over v} + {1 \over {30}} = {1 \over {20}}

v = +60 cm If image position does not change even when mirror is removed it means image formed by lens is formed at centre of curvature of spherical mirror.

Radius of curvature of mirror = 80 – 60 = 20 cm.

$\Rightarrow$ Focal length of mirror f = 10 cm for virtual image, object is to be kept between focus and pole.

$\Rightarrow$ Maximum distance of object from spherical mirror for which virtual image is formed, is 10 cm.

Q45

An upright object is placed at a distance of 40 cm in front of a convergent lens of focal length 20 cm. A convergent mirror of focal length 10 cm is placed at a distance of 60 cm on the other side of the lens. The position and size of the final image will be :

A 20 cm from the convergent mirror, same size as the object

B 40 cm from the convergent mirror, same size as the object

C 40 cm from the convergent lens, same size as the object

D 20 cm from the convergent mirror, twice the size of the object

Correct Answer

Option C

Solution

In given system of lens and mirror, position of object O in front of lens is at a distance 2f. i.e. u = 2f = 40 cm So, image (I1) formed is real, inverted and at a distance, v = 2f = 2 $\times$ 20 = 40 cm, (behind lens) magnification,

{m_1} = {v \over u} = {{40} \over {40}} = 1

Thus, size of image is same as that of that of object.

This image (I1) acts like a real object for mirror.

As object distance for mirror is u = C = 2f = $-$ 20 cm where, C = centre of curvature.

So, image (I2) formed by mirror is at 2f.

$\therefore$ For mirror v = 2f = 2( $-$ 10) = $-$ 20 cm Magnification,

{m_2} = - {v \over u} = - {{( - 20)} \over {( - 20)}} = - 1

Thus, image size is same as that of object.

The image I2 formed by the mirror will act like an object for lens.

As the object is at 2f distance from lens, so image (I3) will be formed at a distance 2f or 40 cm.

Thus, magnification,

{m_3} = {v \over u} = {{40} \over {40}} = 1

So, final magnification,

m = {m_1} \times {m_2} \times {m_3} = - 1

Hence, final image (I3) is real, inverted of same size as that of object and coinciding with object.

Q46

A plano-convex lens (focal length f2, refractive index

\mu

2, radius of curvature R) fits exactly into a plano-concave lens (focal length f1, refractive index

\mu

1, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be :

A f1 + f2

B f1

-

f2

C

{R \over {{\mu _2} - {\mu _1}}}

D

{{2{f_1}{f_2}} \over {{f_1} + {f_2}}}

Correct Answer

Option C

Solution

{1 \over F} = {1 \over {{f_1}}} + {1 \over {{f_2}}} = {{1 - {\mu _1}} \over R} + {{{\mu _2} - 1} \over R}

Q47

The incident ray, reflected ray and the outward drawn normal are denoted by the unit vectors

\overrightarrow a

,

\overrightarrow b

and

\overrightarrow c

respectively. Then choose the correct relation for these vectors.

A

\overrightarrow b

=

\overrightarrow a

+ 2

\overrightarrow c

B

\overrightarrow b

=

\overrightarrow a

-

2 (

\overrightarrow a

.

\overrightarrow c

)

\overrightarrow c

C

\overrightarrow b

= 2

\overrightarrow a

+

\overrightarrow c

D

\overrightarrow b

=

\overrightarrow a

-

\overrightarrow c

Correct Answer

Option B

Solution

Here

\overrightarrow a = \left| {\overrightarrow a } \right|\sin \theta \widehat i - \left| {\overrightarrow a } \right|\cos \theta \widehat j

As

\overrightarrow a

is an unit vector, so

\left| {\overrightarrow a } \right|

= 1 $\therefore$

\overrightarrow a = \left| {\overrightarrow a } \right|\sin \theta \widehat i - \left| {\overrightarrow a } \right|\cos \theta \widehat j

=

\sin \theta \widehat i - \cos \theta \widehat j

Similarly

\overrightarrow b = \sin \theta \widehat i + \cos \theta \widehat j

and

\overrightarrow c = \widehat j

From option (B),

\overrightarrow a

$-$ 2 (

\overrightarrow a

.

\overrightarrow c

)

\overrightarrow c

=

\sin \theta \widehat i + \cos \theta \widehat j

=

\overrightarrow b

Q48

Given below are two statements : one is labeled as Assertion A and the other is labeled as Reason R. Assertion A : For a simple microscope, the angular size of the object equals the angular size of the image. Reason R : Magnification is achieved as the small object can be kept much closer to the eye than 25 cm and hence it subtends a large angle. In the light of the above statements, choose the most appropriate answer from the options given below :

A A is true but R is false

B A is false but R is true

C Both A and R are true and R is the correct explanation of A

D Both A and R are true but R is NOT the correct explanation of A

Correct Answer

Option C

Solution

The formation of image with simple microscope is shown below. Here,

\theta ' = {h \over {{u_0}}} = {{h'} \over D} = {{h'} \over {25}}

where, D = 25 cm (least distance of distinct vision) Here, $\theta$ ' is same for both object and image, hence Assertion is true.

Magnification,

m = {{\theta '} \over \theta } = {D \over {{u_0}}}

Hence, if u0 < D (25 cm), hence, the value of $\theta$ ' will obtain large. So, option (c) is the correct.

Q49

The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same refractive index? Assume the radii of curvature of the faces of lens are R1 and R2 respectively.

A Zero

B Infinite

C 1

D

{{{R_1}{R_2}} \over {{R_1} - {R_2}}}

Correct Answer

Option B

Solution

Given, initially refractive index (n1) = 1.4 Then it placed in medium of same refractive index. $\therefore$ n2 = 1.4 We know, Focal length

{1 \over f} = \left( {{{{n_1}} \over {{n_2}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

$\Rightarrow$

{1 \over f} = \left( {{{1.4} \over {1.4}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

$\Rightarrow$

{1 \over f} = 0

$\Rightarrow$ f = infinite

Q50

The thickness at the centre of a plane convex lens is 3 mm and the diameter is 6 cm. If the speed of light in the material of the lens is 2

\times

108 ms

-

1. The focal length of the lens is ____________.

A 0.30 cm

B 30 cm

C 15 cm

D 1.5 cm

Correct Answer

Option B

Solution

{R^2} = {3^2} + {(R - 0.3)^2}

$\Rightarrow$

{R^2} = 9 + {R^2} + 0.09 - 2 \times 0.3R

$\Rightarrow$

2 \times 0.3R = 9.09

$\Rightarrow$

R = 15.15

cm

\mu = {C \over V}

=

{{3 \times {{10}^8}} \over {2 \times {{10}^8}}}

= 1.5

{1 \over f} = (1.5 - 1)\left( {{1 \over R}} \right)

f \simeq 30

cm