Question

A transparent sphere of radius R and
refractive index n is kept in air. At what distance
from the surface of the sphere should a point
object be placed on the principal axis so as to
form a real image at the same distance from the
second surface of the sphere?

Answer 1

Answer:

Step by step solution :

STEP

1

:

Equation at the end of step 1

(32t2 - 64t) - 64 = 0

STEP

2

:

Trying to factor by splitting the middle term

2.1 Factoring 9t2-64t-64

The first term is, 9t2 its coefficient is 9 .

The middle term is, -64t its coefficient is -64 .

The last term, "the constant", is -64

Step-1 : Multiply the coefficient of the first term by the constant 9 • -64 = -576

Step-2 : Find two factors of -576 whose sum equals the coefficient of the middle term, which is -64 .

-576 + 1 = -575

-288 + 2 = -286

-192 + 3 = -189

-144 + 4 = -140

-96 + 6 = -90

-72 + 8 = -64 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -72 and 8

9t2 - 72t + 8t - 64

Step-4 : Add up the first 2 terms, pulling out like factors :

9t • (t-8)

Add up the last 2 terms, pulling out common factors :

8 • (t-8)

Step-5 : Add up the four terms of step 4 :

(9t+8) • (t-8)

Which is the desired factorization

Equation at the end of step

2

:

(t - 8) • (9t + 8) = 0

STEP

3

:

Theory - Roots of a product

3.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.