योग्य जोडी लावा :
\[\begin{array}{|c|c|} \hline \textbf{स्तंभ 'अ'} & \textbf{स्तंभ 'ब'} \\ \hline \text{पदार्थ} & \text{अपवर्तनांक} \\ \hline \text{हवा} & \text{1.0003} \\ \hline \end{array}\]
Step 1: संकल्पना समजून घेणे.
अपवर्तनांक (Refractive Index) हा पदार्थामध्ये प्रकाशाचा वेग किती कमी होतो हे दर्शवणारा गुणधर्म आहे. माध्यम जितके घनदाट, तितका त्याचा अपवर्तनांक जास्त असतो.
Step 2: विश्लेषण.
हवा हे सर्वात विरळ माध्यम असल्यामुळे त्यामधील प्रकाशाचा वेग सर्वाधिक असतो. त्यामुळे हवेचा अपवर्तनांक खूप कमी म्हणजेच जवळजवळ 1 असतो. त्याची अचूक किंमत सुमारे 1.0003 आहे.
Step 3: निष्कर्ष.
हवा — अपवर्तनांक 1.0003 ही योग्य जोडी आहे.
A parallel beam of light travelling in air (refractive index \(1.0\)) is incident on a convex spherical glass surface of radius of curvature \(50 \, \text{cm}\). Refractive index of glass is \(1.5\). The rays converge to a point at a distance \(x \, \text{cm}\) from the centre of curvature of the spherical surface. The value of \(x\) 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, \( d = 4/3 \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), 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. 
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, \( d = 4/3 \) cm and \( \theta_1 = \theta_2 = \cos^{-1} \frac{n_2}{2n_1} \), 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. 
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\). 
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\)) 
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.