The objective and eyepiece of an astronomical telescope are double convex lenses with refractive index 1.5. When the telescope is adjusted to infinity, the separation between the two lenses is 16 cm. If the space between the lenses is now filled with water and again telescope is adjusted for infinity, then the present separation between the lenses is

To find the present separation between the lenses when the space between them is filled with water, we need to consider the refractive index of water (let's assume it is 1.33).

We can use the lens formula to calculate the separation between the lenses. The lens formula is given by:

1/f = (n - 1) * (1/R1 - 1/R2)

where:
- f is the focal length of the lens
- n is the refractive index of the medium surrounding the lens (in this case, either air or water)
- R1 and R2 are the radii of curvature of the two lens surfaces

Assuming the lenses are identical, they will have the same focal length and radii of curvature. Let's call the focal length f and the radii of curvature R.

When the telescope is adjusted to infinity in air, the separation between the lenses is given as 16 cm. Let's call it d.

Using the lens formula, we can write the equation as:

1/f = (1.5 - 1) * (1/R - 1/R) [Since both lenses are double convex]

Simplifying the above equation, we get:

1/f = 0

Since the telescope is adjusted to infinity, the focal length is infinite. This means the light rays from distant objects are parallel and converge at the eyepiece.

Now, when the space between the lenses is filled with water, we can calculate the present separation between the lenses. Let's call it d'.

Using the lens formula again, we have:

1/f' = (1.33 - 1) * (1/R - 1/R)

Simplifying the equation, we get:

1/f' = 0.33/R

Since f' is also infinite, the focal length remains unchanged. Therefore, the term 1/f' becomes zero. This implies that the separation between the lenses, d', can be any value as long as the focal length remains infinite.

Hence, the present separation between the lenses when the space between them is filled with water can vary, as long as the focal length remains infinite.

fsumnew=16*(1.5-1)/(1.5-1.33)

= 16(.5/.17)= 47 cm

and because the diatance between the lenses is the sum of the two focal distances, that is the new distance.