Ncert Solutions Physics Class 12th

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New answer posted•a year ago
S
a year ago

9.12 Focal length of the objective lens, fo = 8 mm = 0.8 cm

Focal length of the eyepiece, fe = 2.5 cm

Object distance for the objective lens, uo = - 9.0 mm = -0.9 cm

Least distance of distant vision, d = 25 cm

Image distance of the eyepiece, ve = -d = -25 cm

Object distanc

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New answer posted•a year ago
S
a year ago

9.11 Focal length of the objective lens, f1 = 2.0 cm

Focal length of the eyepiece, f2 = 6.25 cm

Distance between the objective lens and the eyepiece, d = 15 cm

Least distance of distinct vision, d’ = 25 cm

Hence, image distance for the eyepiece, v2 = - 25 cm

Let the object distanc

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New answer posted•a year ago
S
a year ago

9.10 Focal length of the convex lens,  f1 = 30 cm, focal length of the concave lens,  f2 = -20 cm

Let the focal length of the combined lens be = f

The equivalent focal length of a system of two lenses in combined form is given by

1f = 1f1 + 1f2

1f = 130 - 

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New answer posted•a year ago
S
a year ago

9.9 Size of the object, h1 = 3 cm

Object distance, u = - 14 cm

Focal length of the concave lens, f = - 21 cm

Image distance = v

According to lens formula

1v - 1u = 1f or 1v - 1-14 = -121

1v = -114-121 or v = -425 = - 8.4 cm

Hence, the image is formed on

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New answer posted•a year ago
S
a year ago

9.8 Object distance, u = 12 cm

Focal length of the convex lens, f = 20 cm

Image distance = v

According to lens formula

1v - 1u = 1f or 1v - 112 = 120

1v = 120+112 or v = 608 = 7.5 cm

Hence,theimageisformed7.5cmawayfromthelens,towardsitsright.

Focal length of the concave lens, f = -16 cm

Image distance = v

Accordi

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New answer posted•a year ago
S
a year ago

9.7 Refractive index of glass, μ = 1.55

Focal length required for the double-convex lenses, f = 20 cm

Let the radius of curvature of one face of the lens be = R1 and the other face be = R2

Let the radius of curvature of the double convex lenses be = R

Then R1=R and R2=-R

Th

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New answer posted•a year ago
S
a year ago

9.6 The angle of minimum deviation, δm = 40 °

Angle of prism, A = 60 °

Let the refractive index of water, =1.33 , and the refractive index of prism material = μ'

The angle of deviation is related to refractive index μ' is given as

μ'=sin?(A+δm)2sin?A2 = sin?(60°+40°)2sin?60°2 = sin?50°sin?30° = 1.532

So

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New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
S
a year ago

11.35 Room temperature, T = 27 ?  = 300 K

Atmospheric pressure, P = 1 atm = 1.01 × 10 5  Pa

Atomic weight of helium atom = 4

Avogadro’s number, N A  = 6.023 × 10 23

Boltzmann’s constant, k = 1.38 × 10 - 23  J m o l - 1 K - 1

Planck’s constant, h = 6.626 × 10 - 34  Js

Average energy of a gas at temperature T is given as:

E =

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