Ray Optics And Optical Instruments

Here, , rad,

Now, where D is the diameter of objective.

Area of the bright patch on the water surface is seen due to total internal reflection of light.
Apply Snell's law at the interface of water and air.


or , or,
or, or,
Area of bright patch on the water surface

Answer (a) to the horizontal

The focal length of the plano-convex lens is 20 cm with the plane surface being silvered.
So, the net power of the lens is given by

Thus, the net focal length is

Apply Snell;s law at the interface,
or
Here,
So,

Refracting angle of prism,

Using Snell's law at point ,


or,

Similar to object at centre of a sphere

The image formed by the concave mirror is enlarged, erect and virtual when the object is placed between pole and focus as shown in figure.
In all other given positions of the object, the image is real.

Here, angle of prism,
Refractive index of prism,
Refractive index of water,
Angle of minimum deviation,
As

Here, for real image, or

Also, so
Using mirror formuIa,

Step 1: Understanding the situation

The problem involves a plane mirror that is silvered on its back surface, making it a "silvered mirror." The mirror has a thickness of 3 cm, and the refractive index . A point object is placed 9 cm from the front surface of the mirror. Since the mirror is silvered on the back surface, the light travels through the glass, reflects off the back surface, and then exits through the front surface. We need to find the position of the image formed by this setup.

Step 2: Effective thickness of the glass

When light enters the glass, it first travels a distance equal to the thickness of the glass (3 cm), and since the refractive index is greater than 1, the apparent distance traveled by light is shortened. The effective thickness of the glass for light is given by: So, the light appears to travel only 2 cm within the glass due to the refractive index.

Step 3: Image formation by the plane mirror

For a plane mirror, the image is formed at the same distance behind the mirror as the object is in front of it, except for the additional path the light travels through the glass. The object is 9 cm from the front surface, so the effective object distance from the mirror's silvered back surface becomes: This is the distance from the back surface where the image would form (virtual image due to the silvering).

Step 4: Find the final position of the image

The image formed is behind the mirror, at a distance of 11 cm from the back surface. Since the distance from the front surface to the back surface is 3 cm, the final position of the image from the front surface is:

Final Answer:

The position of the brightest image from the front surface of the mirror is .

The magnification produced by a telescope is given by the formula: where: - is the image distance for the eyepiece, - is the object distance for the eyepiece. The image formed by the objective lens serves as the object for the eyepiece. For the eyepiece, the image is formed at the least distance of distinct vision (i.e., ). The object distance for the eyepiece is given by: where is the focal length of the eyepiece. Substituting the known values: Thus: Now, the magnification produced by the telescope is the product of the magnifications of the objective and the eyepiece: The magnification due to the objective is: where: - is the image distance for the objective lens, - is the object distance for the objective. For the objective, using the lens formula , where , we find: Thus: So, the magnification due to the objective is: Finally, the total magnification produced by the telescope is the product of these two magnifications:
Thus, the magnification produced by the telescope is .