In an experiment to measure the focal length (f) of a convex lens, the magnitude of object distance (x) and the image distance (y) are measured with reference to the focal point of the lens. The y-x plot is shown in figure.
The focal length of the lens is_____cm.

From the lens formula:
\[\frac{1}{f} = \frac{1}{x} + \frac{1}{y}.\]
The plot shows the relationship between y (image distance) and x (object distance). For any specific point on the graph, the values of x and y can be substituted into the lens formula to calculate f.
Alternatively, the product of x and y at corresponding points can be used to directly calculate the focal length. From the lens equation, we know:
\[x \cdot y = f^2.\]
From the graph: - At point C, x = 20 cm and y = 20 cm.
Using \(x \cdot y = f^2\):
\[f^2 = 20 \cdot 20 = 400.\]
Taking the square root:
\[f = \sqrt{400} = 20 \text{ cm}.\]
Final Answer: The focal length of the lens is:
\[\boxed{20 \text{ cm}}.\]
The problem provides a plot of the magnitude of image distance (y) versus the magnitude of object distance (x) for a convex lens. Both distances are measured with reference to the focal points of the lens. We need to determine the focal length (f) of the lens from this plot.
When the object distance (x) and image distance (y) for a lens are measured from the respective focal points, their relationship with the focal length (f) is given by Newton's Lens Formula. The formula is:
\[ xy = f^2 \]This equation describes a hyperbolic relationship between x and y. To find the focal length, we can select a coordinate pair (x, y) from the experimental curve shown in the plot and substitute it into this formula.
Step 1: Identify the governing equation. According to the problem description, the distances x and y are measured from the focal points. Therefore, we use Newton's Lens Formula:
\[ xy = f^2 \]Step 2: Extract a data point from the given y-x plot. The curve represents the experimental data. Several points (A, B, C, D, E) are shown on the curve. Point C is clearly marked with dashed lines indicating its coordinates on the x and y axes.
From the graph at point C:
\[ x = 20 \, \text{cm} \] \[ y = 20 \, \text{cm} \]Step 3: Substitute the coordinates of point C into Newton's Lens Formula to find the value of \(f^2\).
\[ f^2 = x \times y \] \[ f^2 = (20 \, \text{cm}) \times (20 \, \text{cm}) \] \[ f^2 = 400 \, \text{cm}^2 \]Step 4: Calculate the focal length (f) by taking the square root of \(f^2\).
\[ f = \sqrt{400 \, \text{cm}^2} \] \[ f = 20 \, \text{cm} \]The focal length of the convex lens is 20 cm.
Match List-I with List-II for the index of refraction for yellow light of sodium (589 nm)
| LIST-I (Materials) | LIST-II (Refractive Indices) | ||
|---|---|---|---|
| A. | Ice | I. | 1.309 |
| B. | Rock salt (NaCl) | II. | 1.460 |
| C. | CCl₄ | III. | 1.544 |
| D. | Diamond | IV. | 2.417 |
Choose the correct answer from the options given below:
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: 
Two light beams fall on a transparent material block at point 1 and 2 with angle \( \theta_1 \) and \( \theta_2 \), respectively, as shown in the figure. After refraction, the beams intersect at point 3 which is exactly on the interface at the other end of the block. Given: the distance between 1 and 2, \( d = \frac{4}{3} \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \left( \frac{n_2}{2n_1} \right) \), where \( n_2 \) is the refractive index of the block and \( n_1 \) is the refractive index of the outside medium, then the thickness of the block is …….. cm.

Let \( y^2 = 12x \) be the parabola and \( S \) its focus. Let \( PQ \) be a focal chord of the parabola such that \( (SP)(SQ) = \frac{147}{4} \). Let \( C \) be the circle described by taking \( PQ \) as a diameter. If the equation of the circle \( C \) is: \[ 64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta, \] then \( \beta - \alpha \) is equal to: