Ncert Solutions Physics Class 12th

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New answer posted•a year ago
S
a year ago

9.22 The incident, refracted and emergent rays associated with a glass prism ABC is shown in the adjoined figure.

Angle of the prism, ∠A = 60 °

Refractive index of the prism, μ = 1.524

Let ∠i1 be the incident angle, ∠r1 be the refracted angle, ∠r2 be the incide

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

9.20 Distance between the object and the image ( screen), D = 90 cm

Distance between two locations of the convex lens, d = 20 cm

Let the focal length of the lens be = f

Focal length is related to D and d as:

f = D2-d24D = 902-2024*90 = 21.39 cm

Therefore, the focal length of the convex lens is 21.39 cm

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New answer posted•a year ago
S
a year ago

9.19 Distance between the object and the image, d = 3 m

Let maximum focal length be fmax

For real image, the maximum focal length is given as:

fmax=d4 = 34 = 0.75 m

Hence, for this purpose, the maximum possible focal length is 0.75 m

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

9.17 Refractive index of the glass fibre, μ1 = 1.68

Refractive index of outer covering of the pipe, μ2 = 1.44

Given:-

Angle of incidence = i, Angle of refraction = r, Angle of incidence at the interface = i’

The refractive index of the inner core-outer core interface, μ is given as

μ=μ2μ1&n

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New answer posted•a year ago
S
a year ago

9.16 The actual depth of the pin, d = 15 cm

Let the apparent depth of the pin be = d’

Refractive index of the glass, μ = 1.5

Ratio of actual depth to the apparent depth is equal to the refractive index of the glass. i.e.

μ=dd' or

d’ = dμ = 151.5 = 10

The distance at which the pin app

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

9.14 Focal length of the objective lens, fo = 15 m = 1500 cm

Focal length of the eyepiece, fe = 1.0 cm

The angular magnification of the telescope is given by,

m = fofe = 15001 = 1500

Hence the angular magnification of the telescope is 1500

Diameter of the moon, dm = 3.4

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New answer posted•a year ago
S
a year ago

9.13 Focal length of the objective lens,  fo = 144 cm

Focal length of the eyepiece,  fe = 6.0 cm

The magnifying power of the telescope, m = fofe = 1446 = 24

The separation between the eyepiece and objective lens = fe+fo = 6 + 144 = 150 cm

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