Geometrical Optics — Page 4 | JEE Physics

Q31

A point like object is placed at a distance of 1 m in front of a convex lens of focal length 0.5 m. A plane mirror is placed at a distance of 2 m behind the lens. The position and nature of the final image formed by the system is :

A 2.6 m from the mirror, real

B 1 m from the mirror, real

C 2.6 m from the mirror, virtual

D 1 m from the mirror, virtual

Correct Answer

Option A

Solution

Object is at 2f.

So image will also be at 2f.

(I1).

Image(I2) of I1 will be 1 m behind mirror.

Now I2 will be object for lens. $\therefore$ u = -3 m f = + 0.5 m

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

=

{1 \over {0.5}} + {1 \over { - 3}}

$\Rightarrow$ v =

{3 \over 5}

= 0.6 m So total distance from mirror = 2 + 0.6 = 2.6 m and real image.

Q32

A scientist is observing a bacteria through a compound microscope. For better analysis and to improve its resolving power he should. (Select the best option)

A Decrease the focal length of the eye piece.

B Increase the wave length of the light

C Increase the refractive index of the medium between the object and objective lens

D Decrease the diameter of the objective lens

Correct Answer

Option C

Solution

Resolving power of microscope

= \left( {{{2n\sin \theta } \over \lambda }} \right)

n\sin \theta

= Numerical aperture

n

is the refractive index of medium.

Q33

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

Correct Answer

Option D

Solution

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

Q34

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

Correct Answer

Option C

Solution

$\therefore$

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

Q35

A green light is incident from the water to the air - water interface at the critical angle

\left( \theta \right)

. Select the correct statement.

A The entire spectrum of visible light will come out of the water at an angle of

{90^ \circ }

to the normal.

B The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.

C The spectrum of visible light whose frequency is more than that of green light will come out to the air medium.

D The entire spectrum of visible light will come out of the water at various angles to the normal.

Correct Answer

Option B

Solution

For critical angle

{\theta _c},

\sin {\theta _c} = {1 \over \mu }

For greater wavelength or lesser frequency $\mu$ is less.

So, critical angle would be more, So, they will not suffer reflection and come out at angles less then

{90^ \circ }.

Q36

A thin convex lens made from crown glass

\left( {\mu = {3 \over 2}} \right)

has focal length

f

. When it is measured in two different liquids having refractive indices

{4 \over 3}

and

{5 \over 3},

it has the focal lengths

{f_1}

and

{f_2}

respectively. The correct relation between the focal lengths is :

A

{f_1} = {f_2} < f

B

{f_1} > f

and

{f_2}

becomes negative

C

{f_2} > f

and

{f_1}

becomes negative

D

{f_1}\,

and

{f_2}\,

both become negative

Correct Answer

Option B

Solution

By Lens maker's formula for convex lens

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

for,

\mu {L_1} = {4 \over 3},{f_1} = 4R

for

\mu {L_2} = {5 \over 3},{f_2} = - 5R

\Rightarrow {f_2} = \left( - \right)ve

Q37

To determine refractive index of glass slab using a travelling microscope, minimum number of readings required are :

A Two

B Three

C Four

D Five

Correct Answer

Option B

Solution

The refractive index is

\mu = {{{\mathop{\rm Real}\nolimits} \,depth} \over {Apparent\,depth}} = {{{\mathop{\rm Reading}\nolimits} \,3 - Reading\,1} \over {{\mathop{\rm Reading}\nolimits} \,3 - Reading\,2}}

Therefore, the minimum of three readings are required.

Q38

An observer looks at a distant tree of height

10

m

with a telescope of magnifying power of

20.

To the observer the tree appears:

A

20

times taller

B

20

times nearer

C

10

times taller

D

10

times nearer

Correct Answer

Option B

Solution

A telescope magnifies by making the object appearing closer.

Q39

In an experiment for determination of refractive index of glass of a prism by

i - \delta ,

plot it was found thata ray incident at angle

{35^ \circ }

, suffers a deviation of

{40^ \circ }

and that it emerges at angle

{79^ \circ }.

In that case which of the following is closest to the maximum possible value of the refractive index?

A

1.7

B

1.8

C

1.5

D

1.6

Correct Answer

Option C

Solution

We know that

i + e - A = \delta

{35^ \circ } + {79^ \circ } - A = {40^ \circ }

$\therefore$

A = {74^ \circ }

But

\mu = {{\sin \left( {{{A + {\delta _m}} \over 2}} \right)} \over {\sin A/2}} = {{\sin \left( {{{74 + \delta } \over 2}} \right)} \over {\sin {{74} \over 2}}}

= {5 \over 3}\sin \left( {{{37}^ \circ } + {{{\delta _m}} \over 2}} \right)

{\mu _{\max }}\,

can be

{5 \over 3}.

That is

{\mu _{\max }}

is less than

{5 \over 3} = 1.67

But

{\delta _m}

will be less than

{40^ \circ }

so

\mu < {5 \over 3}\sin \,{57^ \circ } < {5 \over 3}\,\,

\sin {60^ \circ } \Rightarrow \mu = 1.45

Q40

In an experiment a convex lens of focal length 15 cm is placed coaxially on an optical bench in front of a convex mirror at a distance of 5 cm from it. It is found that an object and its image coincide, if the object is placed at a distance of 20 cm from the lens. The focal length of the convex mirror is :

A 27.5 cm

B 20.0 cm

C 25.0 cm

D 30.5 cm

Correct Answer

Option A

Solution

The given optical situation is depicted in the following ray diagram: In this case, the image forms 55 cm behind the convex mirror and then the reflection takes place due to the mirror image that forms at a distance (v) behind the mirror.

Now, this image acts as an object for lens and the final image forms at a distance of 20 m from the lens if the focal length of convex mirror is as follows:

{{55} \over 2} = 27.5

cm