मराठी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान इयत्ता ११

A Candle Flame 1.6 Cm High is Imaged in a Ball Bearing of Diameter 0.4 Cm. If the Ball Bearing is 20 Cm Away from the Flame, Find the Location and the Height of the Image. - Physics

Advertisements
Advertisements

प्रश्न

A candle flame 1.6 cm high is imaged in a ball bearing of diameter 0.4 cm. If the ball bearing is 20 cm away from the flame, find the location and the height of the image.

बेरीज
Advertisements

उत्तर

Given,
Height (h1) of the candle flame taken as object AB = 1.6 cm
Diameter of the ball bearing (d) = 0.4 cm
So, radius = 0.2 cm
Distance of object, u = 20 cm

Using mirror formula,
\[\frac{1}{v} + \frac{1}{u} = \frac{2}{R}\]
Putting the values according to sign conventions, we get,

\[\frac{1}{( - 20)} + \frac{1}{v} = \frac{2}{0 . 2}\] 

\[ \Rightarrow   \frac{1}{v} = \frac{1}{20} + 10\] 

\[ \Rightarrow   v = 0 . 1  cm  or  1 . 0  \text{ mm  inside  the  ball  bearing . }\] 

\[Magnification = m\] 

\[ = \frac{A'B'}{AB} =  - \frac{v}{u} = m\] 

\[ = \frac{A'B'}{200} = \frac{1}{200}\] 

\[\Rightarrow A'B'=\frac{AB}{200}=+\frac{1 . 6}{200} =  + 0 . 08  \text{ cm }  (+0008\text{ cm })\]

Hence, the distance of the image is 1 cm.
Height of the image is 0.008 cm.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 18: Geometrical Optics - Exercise [पृष्ठ ४१२]

APPEARS IN

एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 18 Geometrical Optics
Exercise | Q 5 | पृष्ठ ४१२

संबंधित प्रश्‍न

Why can’t we see clearly through fog?


A concave mirror having a radius of curvature 40 cm is placed in front of an illuminated point source at a distance of 30 cm from it. Find the location of the image.


A 3 cm tall object is placed at a distance of 7.5 cm from a convex mirror of focal length 6 cm. Find the location, size and nature of the image.


A converging mirror M1, a point source S and a diverging mirror M2 are arranged as shown in figure. The source is placed at a distance of 30 cm from M1. The focal length of each of the mirrors is 20 cm. Consider only the images formed by a maximum of two reflections. It is found that one image is formed on the source itself. (a) Find the distance between the two mirrors. (b) Find the location of the image formed by the single reflection from M2.


Locate the image of the point P as seen by the eye in the figure.


k transparent slabs are arranged one over another. The refractive indices of the slabs are μ1, μ2, μ3, ... μk and the thicknesses are t1 t2t3, ... tk. An object is seen through this combination with nearly perpendicular light. Find the equivalent refractive index of the system which will allow the image to be formed at the same place.


An optical fibre (μ = 1.72) is surrounded by a glass coating (μ = 1.50). Find the critical angle for total internal reflection at the fibre-glass interface.


Light falls from glass (μ = 1.5) to air. Find the angle of incidence for which the angle of deviation is 90°.


A point source is placed at a depth h below the surface of water (refractive index = μ). (a) Show that light escapes through a circular area on the water surface with its centre directly above the point source. (b) Find the angle subtended by a radius of the area on the source.


The diameter of the sun is 1.4 × 109 m and its distance from the earth is 1.5 × 1011 m. Find the radius of the image of the sun formed by a lens of focal length 20 cm.


Explain: ‘How is a rainbow formed’?


Answer the following question in detail.

Explain the formation of a primary rainbow. For which angular range with the horizontal is it visible?


State any one difference between a primary rainbow and a secondary rainbow.


Case study: Mirage in deserts

To a distant observer, the light appears to be coming from somewhere below the ground. The observer naturally assumes that light is being reflected from the ground, say, by a pool of water near the tall object.

Such inverted images of distant tall objects cause an optical illusion to the observer. This phenomenon is called mirage. This type of mirage is especially common in hot deserts.

Based on the above facts, answer the following question :

In an optical fibre, if n1 and n2 are the refractive indices of the core and cladding, then which among the following, would be a correct equation? 


Case study: Mirage in deserts 

To a distant observer, the light appears to be coming from somewhere below the ground. The observer naturally assumes that light is being reflected from the ground, say, by a pool of water near the tall object.

Such inverted images of distant tall objects cause an optical illusion to the observer. This phenomenon is called mirage. This type of mirage is especially common in hot deserts.

Based on the above facts, answer the following question:

A diamond is immersed in such a liquid which has its refractive index with respect to air as greater than the refractive index of water with respect to air. Then the critical angle of diamond-liquid interface as compared to critical angle of diamond-water interface will


Case study: Mirage in deserts

To a distant observer, the light appears to be coming from somewhere below the ground. The observer naturally assumes that light is being reflected from the ground, say, by a pool of water near the tall object.

Such inverted images of distant tall objects cause an optical illusion to the observer. This phenomenon is called mirage. This type of mirage is especially common in hot deserts.

Based on the above facts, answer the following question:

The following figure shows a cross-section of a ‘light pipe’ made of a glass fiber of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for the following phenomena to occur.


The sky would appear red instead of blue if


A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is ______.


A passenger in an aeroplane shall ______.


Between the primary and secondary rainbows, there is a dark band known as Alexandar’s dark band. This is because ______.

  1. light scattered into this region interfere destructively.
  2. there is no light scattered into this region.
  3. light is absorbed in this region.
  4. angle made at the eye by the scattered rays with respect to the incident light of the sun lies between approximately 42° and 50°.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×