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Ray Optics and Optical Instruments

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3 months ago

Ray Optics and Optical Instruments Class 12th

The X-Y plane be taken as the boundary between two transparent media M₁ and M₂. M₁ in $z \geq 0$ has a refractive index of $\sqrt[]{2}$ and M₂ with Z < 0 has refractive index of $\sqrt[]{3} .$ A ray of light travelling in M₁ along the direction given by the vector $\vec{P} = 4 \sqrt[]{3} \hat{i} - 3 \sqrt[]{3} \hat{j} - 5 \hat{k},$ is incident on the plane of separation. The value of difference between the angle of incident in M₁ and the angle of refraction in M₂ will be_______ degree.

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Vishal Baghel

Contributor-Level 10

For angle of incidence 'i' :-

cos I = $\frac{5}{\sqrt[]{48 + 27 + 25}} = \frac{5}{\sqrt[]{100}}$

=> I = 60°

Using snell's law :-

$μ_{1} s i n i = μ_{2} s i n r$

sin r = $\frac{\sqrt[]{2}}{\sqrt[]{3}} s i n 60 ° = \sqrt[]{\frac{2}{3}} * \frac{\sqrt[]{3}}{2} = \frac{1}{\sqrt[]{2}}$

$∴ r = 45 °$

So, difference, I – r = 60° - 45² = 15°

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3 months ago

Ray Optics and Optical Instruments Class 12th

A ray of light is incident at an angle of incidence $60 °$ on the glass slab of refractive index $\sqrt[]{3} .$ After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is $4 \sqrt[]{3}$ cm. The thickness of the glass slab is _______cm.

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Payal Gupta

Contributor-Level 10

$d = 4 \sqrt[]{3}$ cm (Lateral shift)

By Snell's law

$μ_{a i r} s i n 60 ° = μ_{g} s i n θ$

$\Rightarrow 1 * \frac{\sqrt[]{3}}{2} = \sqrt[]{3} s i n θ$

$s i n θ = \frac{1}{2}$

= 30°

$t a n 30 ° = \frac{d}{t} = \frac{4 \sqrt[]{3}}{t}$

$\Rightarrow \frac{1}{\sqrt[]{3}} = \frac{4 \sqrt[]{3}}{t}$

t = 12 cm

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3 months ago

Ray Optics and Optical Instruments Class 12th

Two identical thin biconvex lenses of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is ________ cm.

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Vishal Baghel

Contributor-Level 10

f₁ = 15 cm

$P_{1} = \frac{1}{15} * 100 = \frac{100}{15} D$

$P_{3} = \frac{100}{15} D$

$\frac{1}{f^{2}} = (μ_{2} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$= (1.25 - 1) * \frac{- 2}{R}$

$\frac{1}{f_{2}} = \frac{0.25 * - 2}{15}$

$\frac{1}{f_{2}} = \frac{- 1}{30} c m^{- 1}$

$\begin{array}{l} p_{2} = \frac{1}{f_{2}} = - \frac{100}{30} \\ P_{R} = P_{1} + P_{2} + P_{3} \\ P_{R} = \frac{100}{15} - \frac{100}{30} + \frac{100}{15} \\ P_{R} = 10 D \end{array}$

$P_{R} = \frac{1}{f_{R}}$

$f_{R} = \frac{1}{10} * 100 c m$

$f_{R} = 10 c m$

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3 months ago

Ray Optics and Optical Instruments Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A light ray is incident, at an incident angle q₁, on the system of two plane mirrors M₁ and M₂ having an inclination angle 75° between them (as shown in figure). After reflecting from mirror M₁ it gets reflected back by the mirror M₂ with an angle of reflection 30°. The total deviation of the ray will be ___________ degree.

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Payal Gupta

Contributor-Level 10

$δ_{t o t a l} = 360 ° - 2 θ$

= 360 – 2 * 75° = 210°

Total deviation = $δ = 150 °$

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Ray Optics and Optical Instruments physics ncert exemplar solutions class 12th chapter two

A small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt[]{7} m .$ The refractive index of water is $\frac{4}{3} .$ The area of the surface of water through which light from the bulb can emerge out is xπm². The value of x is_________.

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Vishal Baghel

Contributor-Level 10

by snell's law

$\frac{4}{3} s i n θ_{C} = μ_{1} s i n 90 °$

$s i n θ_{C} = \frac{3}{4} = \frac{r}{\sqrt[]{r^{2} + {(\sqrt[]{7})}^{2}}}$

$\frac{3}{4} = \frac{r}{\sqrt[]{r^{2} + 7}}$

Squaring both side

$\frac{9}{16} = \frac{r^{2}}{r^{2} + 7}$

$A = r^{2} = π * 9 = 9 π$

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Ray Optics and Optical Instruments physics ncert exemplar solutions class 12th chapter two

For a specific wavelength 670nm of light coming from a galaxy moving with velocity v, the observed wavelength is 670.7nm.

The value of v is:

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Vishal Baghel

Contributor-Level 10

$\frac{Δ λ}{λ} = \frac{v}{c}$

$v = \frac{Δ λ}{λ} C$

= $\frac{0.7 * 1 0^{- 9}}{670 * 1 0^{- 9}} * 3 * 1 0^{8}$

= $3.13 * 1 0^{5}$

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3 months ago

Ray Optics and Optical Instruments Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The efficiency of a Carnot's engine, working between stem point and ice point, will be

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Payal Gupta

Contributor-Level 10

Steam point (T₁) = 100°C = 273 + 100 = 373 K

Ice point (T₂) = 0°C = 273 K

$η = (1 - \frac{T_{2}}{T_{1}}) * 100 = (1 - \frac{273}{373}) * 100 = 26.81 %$

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3 months ago

Ray Optics and Optical Instruments Physics NCERT Exemplar Solutions Class 12th Chapter Nine

A parallel beam of light is allowed to afall on a transparent spherical globe of diameter 30cm and refractive index 1.5. The distance from the centre of the globe at which the beam of light can converge is_________ mm.

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alok kumar singh

Contributor-Level 10

When object is at infinity

u = $\infty, μ_{1} = 1, R = 15 c m$

v = ? $μ_{2} = 1.5$

$\frac{- μ_{1}}{u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{- 1}{\infty} + \frac{1.5}{v} =$ $\frac{1.5 - 1}{15}$

$\frac{1.5}{v} =$ $\frac{0.5}{15}$

v =

45cm

Now for Refraction through C2

u = + 15cm

v = ?

$μ_{1} = 1.5$

$μ_{2} = 1$

R = 15

$\frac{- μ_{1}}{u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{- 1.5}{15} + \frac{1}{v} = \frac{1 - 1.5}{- 15}$

$\frac{- 1}{10} + \frac{1}{v} = \frac{1}{30}$

$\frac{1}{v} = \frac{1}{30} + \frac{1}{10} = \frac{4}{10}$

$v = \frac{30}{4} = + 7.5 c m$

from centre = 15 + 7.5 = 22.5 cm = 225mm

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Ray Optics and Optical Instruments Physics NCERT Exemplar Solutions Class 12th Chapter Nine

The graph between $\frac{1}{u} a n d \frac{1}{v}$ for a thin convex lens in order to determine its focal length is plotted as shown in the figure. The refractive index of lens is 1.5 and its both the surfaces have same radius of curvature R. The value of R will be _________cm. (where u = object distance, v = image distance)

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Vishal Baghel

Contributor-Level 10

For point B, $\frac{1}{u} = - 0.10 c m^{- 1}, \frac{1}{v} = 0$

Thus u = 10 cm, v = $\infty$

i.e., f = 10 cm

$\frac{1}{10} = (1.5 - 1) (\frac{2}{R}) \Rightarrow \frac{1}{R} = \frac{1}{10}$

R = 10 cm

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3 months ago

Ray Optics and Optical Instruments Physics NCERT Exemplar Solutions Class 12th Chapter Nine

In the given figure, the face AC of the equilateral prism is immersed in a liquid of refractive index 'n'. For incident angle $60 °$ at the side AC, the refracteel light beam just grazes along face AC. The refractive index of the liquid $n = \frac{\sqrt[]{x}}{4} .$

The value of x is_______.

(Given refractive index of glass = 1.5)

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alok kumar singh

Contributor-Level 10

1.5 * sin 60 = R * sin 90

$\Rightarrow 1.5 * \frac{\sqrt[]{3}}{2} = n = \frac{\sqrt[]{x}}{4}$

x = 27

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