Class 11-science H C VERMA Solutions Physics Chapter 19 - Optical Instruments

For A,

For B,

For C,

For D,

Now, A > B > D > C

Thus, decreasing order is A, B, D, C.

To get maximum angular magnification

v = - D = -25cm

and by formula

Object distance is 8.1cm away from lens.

(a) Magnification of simple microscope is given as:

f = 12.5cm

(b) Magnification of power is given as :

M =2

Maximum angular magnification is given as :

m = 2

Magnification of simple microscope is given as :

f = 5cm

Now, for relaxed farsighted magnifying power is given as :

m = 8

Focal length of objective lens is given as:

fo = 1/25 = "4cm"

Focal length of eyepiece lens is given as:

fe = 1/5 = "20cm"

Now using lens formula for eyepiece lens,

Also, vo = 30 - ue= 30 - 11.11

vo = 18.89cm

Now using lens formula for objective lens,

uo = -5.07cm

Now maximum magnifying power is given as:

m = 8.376

By applying lens formula,

u_e = -4.8cm

(1) If separation is 9.8cm then, v_o =9.8 - 4.8 = 5cm

By applying lens formula,

u_o = -1.25cm

and magnifying power

m = 20

(2) If separation is 11.8cm then, v_o = 11.8 - 4.8 = 7cm

u_o =-1.16cm

And magnifying power,

m = 30

Thus range of magnifying power is 20 to30.

From lens formula,

u_e = -50/7cm

Image distance v_o = 20 - u_e

v_o = 90/7cm

Now applying lens formula on objective lens,

u_o = -90/11cm

Maximum magnification power

m = 5.5

And minimum separation is given as:

0.22/m = 0.22/0.5 = 0.04mm

Magnifying power is given as,

2v_o - 4f_o = 1……(1)

Also,

tube length =v_o + f_o

6.5 = v_o + f_o…….(2)

Solving equation 1 and 2,

v_o = 4.5cm

f_e = 2cm

Here we get figure as below,

By using lens formula (objective)

vo = -1cm

By using lens formula (eyespiece)

uo = -6cm

Now from above figure,

Separation = u_o-v_o= 6-1 = 5cm

Focal length of objective and eyepiece is given as :

fo = 1/25 =4cm, fe= 1/20 = 5cm

(a) fo < fe : it is a microscope.

(b) Angular magnification is given as :

 ………(1)

Now, Vo =25-fe = 25-5 = 20cm

and by lens formula

uo= -5cm

putting values in equation (1)

m = (-20/-5) × 25/5

m = 20

Magnification is given as :

fo = 50fe ……(1)

Also,

L = fo + fe

102 = fo + fe ……(2)

Solving equation 1 and 2 we get,

fo =1cm and fe = 0.02cm

And power is given as :

Po = 1/fo = 1/1 =1D

Pe = 1/fe = 1/0.02 = 50D

Focal length is given as :

fe = 10cm

and

fo = L - fe = 100 - 10

fo = 90cm

Magnifying power ,

m=9

We know that,

L = fo - fe (concave eyepiece lens)

fe = fo - L = 30 - 27

fe = 3cm

For sighted person, lens formula is

f = ⅓ m

Power, P =3D

For near sighted person, formula is

f= -2

And power P  

P = -0.5D

Focal length is given as:

f= -40cm

Lens formula is given as :

v = -40cm

After 10 years,

f= -40cm

u = -50

and by lens formula

v = 200cm

Now to read letter at u = -25cm and v = 200

focal length is:

f = 2/9m

Power P = 1/f = 9/2 = 4.5 D

(a) When lens of eyes is relaxed,

u = ∞ 

v = 0.02m

and

f = 0.02m= 50 D

(b) When lens is in strained position,

u = -0.25

v = 0.02m

and

f = 54D

Power P = 1/f = 54D

Near point = 10cm , u = -10cm

Far point = 100cm , u = -100cm

fn = 1/60cm

P = 1/fn = 60D

Also,

fv = 1/51m

For near sightedness, distance of image from glass,

v = distance of image from eye - distance between glass and eye.

v = 25 - 1= 24 cm

v = 0.24m

and

f = 1/0.24m= 2.4D

P = 4.2D

(a) What we use contact lens is used,

f = 1/4m

P  = 4D

(b) When we use spectacles,

Least distant of distinct vision of glass,

D = 25cm

Focal length of glasses  

Now, without glasses lady will have more value at least distance of distinct vision.

So,

v = -40cm

Focal length of magnifying glass =  

f = 5cm

(a) Maximum magnifying power with glasses is given as :

m = 6

(b) Maximum magnifying power without glasses is given as :

m = 9

As given in question,

v = -40cm, u = -25cm (for left lens of glass)

And, v = -100cm, u = -25cm (for right lens of glass)

(a) For astronomical telescope lady should use right lens having focal length 100/3 cm because eyepiece should have small value of focal length.

(b) With relaxed eye, lady can get magnification as:

m = 2