Class 11-science H C VERMA Solutions Physics Chapter 18 - Geometrical Optics

By using mirror equation,

Hence,

v=-60cm

We know that,

u=-200cm

Now by using mirror formula,

f=-1.44m

There are two possible cases:

(i) u=-u, v=2u, f=-20cm

By using mirror formula,

u=10cm

(ii) u=-u, v=-2u, f=-20cm

By using mirror formula,

u=30cm

Using mirror equation,

As,

{m=}

{0.6=}

u=5cm

Focal length

Now by using mirror formula,

v=0.1cm

and magnification,

h_i = 0.008cm

By using mirror formula ,

And,

Also,

So,

h_i=1.33cm

For PQ,

U=AQ=40cm

v=-13.3cm

Also,

For RS

U=-30cm

v=-15cm

Also,

Now,

Q'S'=AS'-AQ'

=15-13.3

=1.7cm

Thus, total image length is,

3.3+5+1.7=10cm

As we know that,

Now,

f=-87.5cm

By using mirror formula,

Now,

By using mirror formula,

Also,

Let x be height of object above water.

We know,

Both mirror produces one image in following two cases:

(A) When Source is at distance =2f or at center of curvature then image is produce at v=2f+2f=4f

(B) When source is at distance f then rays become parallel after reflecting through one mirror, then it is reflected through other mirror and only one image is formed

V=f + f=2f

By using mirror formula, for reflection,

Now, for 2^nd reflection,

u=60-(30+x)=30-x

Using Mirror equation,

X²+10x-600=0

Solving equation we get,

X=20cm or x=-30cm

Therefore,

Total distance is 20+30=50cm

We know that,

Distance

And time taken,

Total time taken

s

Length of Shadow =0.5+0.5 tan r

Also,

sin r=0.53

and cos r=

=0.85

Length of shadow =0.5(1+tan r)

=0.5(1+0.623)

=81.5cm

According to figure,

x=2.8m

Shift,

Shift due to water,

Shift due to oil,

Adding Shift of water and oil,

=5+4.6

=9.6cm

Apparent depth =40-∆t

=40-9.6

=30.4cm

Shift is given as:

Shift of 0.2cm above P point takes place due to 3 sheets.

Total shift due to 'k ' no. of slabs is:

………… (1)

Where t₁, t₂, t₃…………t_kre thickness and  are refractive index

………………….(2) 

Comparing equation(1) and (2)

When glass piece is inserted inside water then volume of cylindrical water column becomes,

6²×h=800+4²×8

h=25.7cm

Now, shift due to glass block is,

Now, shift due to water is,

Net shift

=2.26+4.44

(A) We know ,

Distance =

Image through mirror

(B) We know that,

Direct image= H+y

Real depth

=

Apparent depth

=

Distance of image==H

By above diagram,

 (1)

Also,

Sin r=1.33 × sin i

Sin r=4/5

Then, cot r=3/4 (2)

Comparing equation (1) and (2),

x=2.25cm

Ration

We know,

As

Let x= distance between P and X

Now

d=x+10

Also,

{} 

According to fig,

And

=24°

Shift = AB sin

=0.62cm

We know,

We know,

And

Hence,  is the largest possible angle.

When light ray travels from glass to air then total internal reflection takes place. In this refracted angle is more than 90° for reflection to occur. Thus, maximum angle is 90°.

We know,

Angle increases from 0° to 40°48' then to 45° angle decreases.

(i) CASE I: When light passes through normal ,

Angle of deviation

(ii) CASE II: When light is incident at critical angle

C=62.73°

Angle of deviation

=90°-c

=90°-62.73°

=37.27°

Range 0 to cos^-1(8/9)

We know that,

Maximum Deviation = 90°-41.8°

=47.2°

Thus, total internal reflection will take place.

In reflection deviation is given as:

D=180°-2i

i=45°

(a)As per figure,

Also,

{}

(b) Angle A= sin^-1(1/)

And also,

x=2.667m

Total radius= 2.667+0.15=2.817m

Angle of minimum deviation is given as :

As we know,

We know that,

As angle is small,

We know that,

As ,

=30°

We know,

We know,

We know that,

And

=2

Therefore,

c=30°

I > c

Therefore, rays reflected internally.

We know,

(a) Image distance from left =-(5-3.5)

=-3.5cm

v=-3cm

Image will be formed at 2cm left from center.

(b) Image distance from right u=-(5+1.5)

=-6.5cm

We know,

Image will be formed at 2.65 cm towards left from center

When rays refract through A then,

v=30cm

Now fro second refraction,

Focused at the surface

We know,

Focused at the center

We know,

This is not possible. Thus, image cannot be focused at the center of the sphere.

We know,

When image is refracted by A

Now when image is refracted at B,

Shift in image is given as:

When refraction takes place from ACB then,

Now, refraction takes of reflected image above so v=f=r/2 and u= -3r/2

Hence,

Therefore, Image is formed at E.

For 1^st refraction,

We know

By using mirror formula,

Now, for 2^nd refraction,

We know,

Air :

f=100cm

Air :

f=300cm

There are four types of lenses:

We know that,

(1) If R1=+ve and R2= -ve

(2) If R1=-ve and R2= -ve

(3) If R1=-ve and R2= +ve

(4) If R1=+ve and R2= +ve

Beam is incident on lens, so,

Now beam is refracted second time,

Therefore, Image is formed at

Image formed when rays are incident from medium with refractive index  is:

(a) We know,

And magnification m=v/u=-490/-9.8=50cm

As m > 1 Then, Image is virtual and erect.

(b) We know,

And magnification m=v/u=510/-9.8=-52.04cm

As m < 1. Then, Image is real and inverted.

In case of projector,

We know,

When object is at A, then

Now, Amplitude of vibration is given as:

Amplitude=2.28cm

We know,

As image formed on left side.

Therefore, Virtual image

 (1)

And

 (2)

 (3)

From equation (1) and (2)

And from eq.(3)

From eq (1) and (3)

We know,

Now,

When the image is tripled then,

We know,

Now,

We know that,

For OB'

For OA'

Now,

A'B'= OB'-OA'

=15-12

A'B'=3cm

We know,

As given,

 (1)

And

 (2) 

Now,

Object is at 15cm from the lens.

We know,

Therefore, For concave mirror,

Now, Using mirror equation for concave mirror,

Now, in case of second refraction,

And

Therefore,

Let x be the object distance.

First using lens formula,

Now,

Now,

Now again it will refract through lens,

We know,

Therefore,

Object distance=15-40/3

=1.67cm

(I) Image formed due to direct transmission through lens.

We know,

(II) Image due to reflection of rays by mirror and then transmission through lens.

In case of lens,

In case of mirror,

Now, distance between mirror and lens is,

We know,

Image will be real and inverted and it will be at a distance of 20cm from mirror and f=10cm.

Hence,

Now again rays will pass through lens, So

Therefore, Image is formed at the place of object. It real and inverted and of same size as object.

We know,

Now, shift of the image due to slab is:

Image is formed at 30+0.33=30.33cm.

For convex lens,

Therefore, For concave lens,

And

Thus, when rays pass through concave lens become parallel

According to fig.

Net focal length is:

F=120cm

Equivalent lens is converging lens.

Now,

Distance of object from diverging lens is:

For d1 120-60= 60cm

For d2 120+90=210cm

For lens I

We know,

Magnification

Image is inverted

Size of image is 10mm.

For lens II

We know,

Magnification

Image is real and erect.

Size of image is 10mm.

We know,

Also

V1=30cm

Now,

Hence

f=10 cm(convex lens)

F=60 cm(concave lens)

Beam will diverge after refracting from both lenses.

1st lens:

2nd lens:

Therefore, Image is formed at distance of 5cm for 1^st convex lens.

(c) Equivalent focal length is given as :

AB =

And BC=

And also u=-()

Now we know,

And velocity

We know,

And velocity V=dv/dt

t < d/V

as we know,

t > d/v

as we know that,

After solving we get,

Absolute velocity is given as

Recoil velocity is given as,

At time 't' position =

And u=kvt

Also,

Velocity is given as:

Thus velocity of separation is given as:

If t=0

In case of equilibrium,

mg=kx

x=mg/k=0.1cm

Thus, mean position =30+0.1=30.1cm

Max. Compression=8

And work done=KE2-KE1

From fig2

Position of B=31.3cm

Amplitude=31.5-30.1=1.4

Position of A=30.1-1.4=28.7cm

For A,

For B,

Therefore, Vibration length= 20.62-19.38=1.24cm

In time t=R/V the mass B must have moved vXR/v=R towards the mirror

For block B

a)

b)

According to given figure:

T-mg+ma-2m=0 (1)

And T-ma=0 (2)

From eq (1) and (2)

2ma-mg-2m=0

2ma=m(g+2)

A=10+2/2=6m/s²

And distance travelled is given as:

S=ut+at²/2

S=0+1/2×6×(0.2)²

S=12cm

Therefore u=-(42-12)

U=-30cm

We know,