Q1. A cylindrical vessel, whose diameter and height both are equal to 30 cm, is placed on horizontal surface and a small particle P is placed in it at a distance of 5 cm from the centre. An eye is placed at a position such that the edge of the bottom is just visible. The particle P is in the plane of drawing. Upto what minimum height should water be poured in the vessel to make the particle P visible?
Q2. An object is placed 20 cm in front of a block of glass 10cm thick and its farther surface is silvered. The image is formed 23.2cm behind the silvered face. Find the refractive index of the glass.
Q3. In a river 2 m deep, a water level measuring post embedded itno the river stands vertically with 1 m of it above the water surface. If the angle of inclination of the sun above the horizon is 30°, calculate the length of the post on the bottom of the river (refractiv index of water = 4/3)
Q4. how long will light take in travelling a distance of 500 metre in water? given that refractiv index for water is 4/3 and the velocity of light in vacuum is 3 x 10(to the power 10) cm/sec. Also calculate equivalent path.

Q5. A ray of sunlight enters a spherical water droplet (refractiv index = 4/3) at an angle of incidence 53° measured with respect to the normal to the surface. It is reflected from the back surface of the droplet and re-enters into air. The angle between the incoming and outgoing ray is [take sin53° = 0.8]
Q6. A cubical vessel with non transparent walls is so loacated that the eye of an observer does not see its bottom, but sees all of the front wall upto bottom corner D. Upto what height water should be poured into ther vessel for the observer to see an object O arranged at a distance of 10 cm from the corner D? The face of vessel is a= 40 cm.

Let's tackle these questions one by one, starting with the first one regarding the cylindrical vessel. The goal is to determine the minimum height of water required to make the particle P visible from a specific viewpoint.

Understanding the Visibility of Particle P

In this scenario, we have a cylindrical vessel with a diameter and height of 30 cm. The particle P is located 5 cm from the center of the base of the cylinder. To make the particle visible from the edge of the bottom, we need to consider the geometry of the situation.

Geometric Considerations

The radius of the cylinder is half of the diameter, which is 15 cm. The distance from the center to the edge of the base is 15 cm. Since the particle is 5 cm from the center, it is located at a distance of:

• Distance from the center to P = 5 cm
• Distance from the center to the edge = 15 cm

To visualize the situation, imagine a right triangle formed by the line of sight from the observer's eye to the particle P, the vertical height of the water, and the horizontal distance from the center of the cylinder to the edge. The observer's eye is at the edge of the bottom, so we need to find the height of the water that allows the line of sight to just touch the water surface before reaching the particle.

Using Similar Triangles

Let’s denote the height of the water as h. The triangle formed by the observer's eye, the water surface, and the particle P can be analyzed using similar triangles. The horizontal distance from the edge to the particle is:

• Horizontal distance = 15 cm - 5 cm = 10 cm

Now, we can set up a proportion using the height of the water (h) and the distances:

• Height of water (h) / Horizontal distance to P (10 cm) = Total height of the cylinder (30 cm) / Radius of the cylinder (15 cm)

This gives us the equation:

h / 10 = 30 / 15

Solving for h:

h / 10 = 2

h = 20 cm

Thus, the minimum height of water that should be poured into the vessel to make the particle P visible is 20 cm.

Moving to the Next Question: Refractive Index of Glass

Now, let's analyze the second question regarding the refractive index of glass. We have an object placed 20 cm in front of a block of glass that is 10 cm thick, with its farther surface silvered. The image is formed 23.2 cm behind the silvered face. We need to find the refractive index of the glass.

Applying the Lens Formula

We can use the lens formula, which relates the object distance (u), image distance (v), and focal length (f) of the system:

1/f = 1/v - 1/u

In this case, the object distance (u) is -20 cm (the negative sign indicates that the object is in front of the lens), and the image distance (v) is +23.2 cm (behind the silvered face). The thickness of the glass (t) is 10 cm.

Calculating the Focal Length

Substituting the values into the lens formula:

1/f = 1/23.2 - 1/(-20)

1/f = 1/23.2 + 1/20

Finding a common denominator (which is 1160):

1/f = (20 + 58) / 1160 = 78 / 1160

Thus, f = 1160 / 78 ≈ 14.87 cm.

Finding the Refractive Index

Now, we can find the refractive index (n) of the glass using the formula:

n = (t / f) + 1

Substituting the values:

n = (10 / 14.87) + 1 ≈ 0.672 + 1 ≈ 1.672.

Therefore, the refractive index of the glass is approximately 1.672.

Next Up: Length of the Post in the River

For the third question, we need to calculate the length of a post embedded in a river that is 2 m deep, with 1 m above the water surface and the sun's angle at 30 degrees.

Understanding the Geometry

The total length of the post (L) can be expressed as:

• L = Length above water + Length below water = 1 m + x

Here, x is the length of the post submerged in the water. The angle of inclination of the sun creates a situation where we can use trigonometry to find x.

Using Snell's Law

We can apply Snell's Law to find the apparent depth of the post. The refractive index of water is given as 4/3. The angle of incidence (i) is 30 degrees, and we can find the angle of refraction (r) using:

n1 * sin(i) = n2 * sin(r)

Substituting the values:

1 * sin(30) = (4/3) * sin(r)

0.5 = (4/3) * sin(r)

sin(r) = 0.5 * (3/4) = 0.375.

Finding the Depth

Now, we can find the actual depth of the post using the relationship between the apparent depth and the real depth:

Apparent depth = Real depth / n

Let’s denote the real depth as D. The apparent depth is 1 m (the length above water), so:

1 = D / (4/3)

D = 1 * (4/3) = 4/3 m ≈ 1.33 m.

Now, we can find the length of the post submerged in the water:

x = D - 1 = 1.33 - 1 = 0.