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To find the angular deviation of a ray of light passing through a parallel-sided glass block, follow these steps: Calculate the angle of refraction at the first surface: Using Snell's Law: \(n_1 \sin(\theta_1) = n_2 \sin(\theta_2)\) where: \(n_1 = 1 \quad (\text{air}), \quad \theta_1 = 30^\circ, \quad n_2 = 1.5 \quad (\text{glass})\) Substituting the values: \(1 \cdot \sin(30^\circ) = 1.5 \cdot \sin(\theta_2) \Rightarrow 0.5 = 1.5 \cdot \sin(\theta_2) \Rightarrow \sin(\theta_2) = \frac{1}{3}\) Thus, \(\theta_2 \approx \arcsin\left(\frac{1}{3}\right) \approx 19.47^\circ\) Angle of incidence at the second surface: Since the block is parallel-sided, the angle of incidence is: \(\theta_3 = \theta_2 \approx 19.47^\circ\) Calculate the angle of refraction at the second surface: Applying Snell's Law again: \(n_2 \sin(\theta_3) = n_1 \sin(\theta_4)\) Substituting values: \(1.5 \cdot \sin(19.47^\circ) = 1 \cdot \sin(\theta_4) \Rightarrow \sin(\theta_4) \approx 0.5 \Rightarrow \theta_4 = \arcsin(0.5) = 30^\circ\) Calculate the angular deviation: The angular deviation \(D\) is given by: \(D = \theta_1 - \theta_4 = 30^\circ - 30^\circ = 0^\circ\) The angular deviation of the emergent ray is \(0^\circ\).

No expanded working has been attached for this question yet. The verified answer is Option B.

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option C.

an upright image is a virtual image and it is expected to be on the same side as the object for a lens but not for a mirror. magnification M = \(\frac{v}{u}\) = \(\frac{v}{2}\) = 3 ⇒⇒ v = 3u ( image is virtual = -3u) using \(\frac{1}{f}\) = \(\frac{1}{u}\) + \(\frac{}{v}\) ⇒⇒ \(\frac{1}{u}\) - \(\frac{1}{3u}\) = \(\frac{1}{12}\) \(\frac{3 -1}{3u}\) = \(\frac{1}{12}\) cross multiplying 24 = 3u = u = 8cm but v = 3u = 24cm virtual image and object are at the side of the lens, Hence, the distance between the object and the image = 24 - 8 = 16cm.

No expanded working has been attached for this question yet. The verified answer is Option B.

v = 4u, if u = 2cm, v = 8cm F = \(\frac{uv}{u + v}\) = \(\frac{2 \times 8}{2 + 8}\) = 1.6cm r = 2F = 2 x 1.6 = 3.2cm

Virtual Image: The image formed is virtual because it cannot be projected on a screen; it appears to be behind the mirror. Erect: The image is upright relative to the object. Enlarged: The image is larger than the object when it is placed between the focal point and the mirror.

No expanded working has been attached for this question yet. The verified answer is Option A.

Total distance = 2 x 160 = 320cm Total time = \(\frac{320}{320}\) = 1s F = \(\frac{1}{1}\) = \(\frac{1}{1}\) = 1Hz

No expanded working has been attached for this question yet. The verified answer is Option C.

From the diagram above: refractive index \(_gn_a\) = \(\frac{1}{_an_g}\) = \(\frac{sin r}{sin i}\) = \(\frac{ sin 30}{ sin 60}\) = \(\frac{1}{0.577}\) = 1.732

Speed: The speed of sound is greater in water than in air. Frequency: The frequency of the sound wave remains constant when it transitions between different mediums. Wavelength: Since the speed increases while the frequency remains constant, the wavelength must increase as well.

\(\lambda\) = \(\frac{v}{f}\) = \(\frac{340}{170}\) = 2m ; for closed pipe L = \(\frac{1}{4}\)\(\lambda\) = \(\frac{1}{4}\) x \(\frac{2}{1}\) = 0.5m = 50cm

u = f