Physics, asked by vai25, 1 year ago

A plano-convex lens of curvature of 30 cm and
refractive index 1.5 produces a real image of an
object kept 90 cm from it. What is the magnification?
A) 4
B) 0.5
C) 1.5
D) 2

Answers

Answered by lolllll6

Answer:

Question:

Two pillars of equal heighi, stand on either side of a roadway which is 150 m wide. From a point on the roadway between the pillars, the elevations of the top of the pillars are 60° and 30°. Find the height of the pillars and the position of the point. Use √3 = 1.73]

Solution:

[ Refer the attachment for figure ]

In ∆ABC

→ tan 30° = AB/BC

→ √3 = AB/BC

→ BC = AB/√3 ...(1)

Similarly, In ∆EDC

→ tan 60° = ED/CD

→ 1/√3 = ED/CD

→ CD = ED√3

Now,

BD = 150 m (given)

We can write BD as BC + CD

So,

→ BD = BC + CD

→ 150 = AB/√3 + ED√3

→ 150 = (AB + ED√3√3)/√3

→ 150 = (AB + 3ED)/√3

→ 150√3 = AB + 3ED

Let AB = ED = h (as both are heights)

→ 150√3 = h + 3h

→ 150√3 = 4h

→ 37.5√3 = h

→ 64.95 = h

So, distance from first point = 64.95/√3 = 37.5 m

Another = 150 - 37.5 = 112.5 m

•°• Height of the pillar is 64.95 m and position of the point from the first point is 64.95/√3 m (37.5) and another is 112.5 m

Question:

In umbrella has 8 ribs which are equally spaced, as shown in the figure. Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:

There are total 8 ribs in the umbrella.

So,

Area of umbrella = 8 (Area of rib)

Area of rib = 1/8 (Area of umbrella)

→ 1/8 (Area of circle)

→ 1/8 × πr²

→ 1/8 × 22/7 × (45)²

→ 1/8 × 22/7 × 2025

→ 22275/28

→ 795.54 cm²

•°• Area between two consecutive ribs of umbrella is 795.54 cm²