When parallel rays from a distant source fall on a spherical glass ball, they refract at the surface, and the rays converge to form an image. Since the ball is a sphere, the incident rays are refracted at both the entry and exit points, forming a real image on the other side of the ball. To find the position of the image, we can apply the formula for the refraction at a spherical surface: \[ \frac{1}{f} = \left( \frac{n - 1}{R} \right) \]
where:
\( f \) is the focal length of the spherical ball,
\( n = 1.5 \) is the refractive index of the glass,
\( R = 15 \, \text{cm} \) is the radius of the spherical ball.
Substitute the values: \[ \frac{1}{f} = \frac{1.5 - 1}{15} = \frac{0.5}{15} = \frac{1}{30} \] So, the focal length \( f = 30 \, \text{cm} \). The ray diagram for this setup is shown below:
The image is formed at a distance of 30 cm from the center of the ball. Therefore, the final image is formed 30 cm away from the center on the opposite side of the incident light. Since the source is far away, the rays converge at this point after refracting through the spherical ball.
Bittu and Chintu were partners in a firm sharing profit and losses in the ratio of 4:3. Their Balance Sheet as at 31st March, 2024 was as
On $1^{\text {st }}$ April, 2024, Diya was admitted in the firm for $\frac{1}{7}$ share in the profits on the following terms:
Prepare Revaluation Account and Partners' Capital Accounts.
(a) Calculate the standard Gibbs energy (\(\Delta G^\circ\)) of the following reaction at 25°C:
\(\text{Au(s) + Ca\(^{2+}\)(1M) $\rightarrow$ Au\(^{3+}\)(1M) + Ca(s)} \)
\(\text{E\(^\circ_{\text{Au}^{3+}/\text{Au}} = +1.5 V, E\)\(^\circ_{\text{Ca}^{2+}/\text{Ca}} = -2.87 V\)}\)
\(\text{1 F} = 96500 C mol^{-1}\)
Define the following:
(i) Cell potential
(ii) Fuel Cell