Ray Optics JEE Main Physics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: (c)

Explanation: In a compound microscope: The objective lens is designed to have a **short (small) focal length**, because high magnification requires a short focal length. To gather more light (improving brightness and resolution), the aperture of this short-focal-length lens is typically **large**. Hence, the correct option is: **(C) small and large.**

Correct Answer: (b)

Explanation: Let's denote the actual depth of the vessel as $d$. Given that the vessel is half-filled with one liquid and the other half with the second liquid, each liquid fills a depth of $\frac{d}{2}$. Let the refractive index of the first liquid be $\mu$ and the refractive index of the second liquid be $2 \mu$. The formula to find the apparent depth due to a single layer of liquid is: $\text{Apparent Depth} = \frac{\text{Actual Depth}}{\text{Refractive Index}}$ First, we find the apparent depth for each segment of the vessel: For the layer with refractive index $\mu$: $\text{Apparent Depth}_1 = \frac{\frac{d}{2}}{\mu}$ For the layer with refractive index $2 \mu$: $\text{Apparent Depth}_2 = \frac{\frac{d}{2}}{2 \mu}$ The total apparent depth is the sum of the apparent depths of both layers: $\text{Total Apparent Depth} = \text{Apparent Depth}_1 + \text{Apparent Depth}_2$ $\text{Total Apparent Depth} = \frac{\frac{d}{2}}{\mu} + \frac{\frac{d}{2}}{2 \mu}$ $\text{Total Apparent Depth} = \frac{d}{2 \mu} + \frac{d}{4 \mu}$ $\text{Total Apparent Depth} = \frac{2d + d}{4 \mu}$ $\text{Total Apparent Depth} = \frac{3d}{4 \mu}$ We are given that the apparent depth of the vessel is $50 \%$ of the actual depth. Therefore: $\frac{3d}{4 \mu} = \frac{d}{2}$ We solve for $\mu$ by isolating it on one side of the equation. Multiply both sides by $4 \mu$: $3d = 2d \mu$ Divide both sides by $2d$: $\mu = \frac{3}{2}$ $\mu = 1.5$ So, the correct option is: **Option B: 1.5**

Correct Answer: (c)

Explanation: For a thin prism, the angle of minimum deviation is $ \delta = (\mu - 1)A $ where $\mu$ is the refractive index of the prism material with respect to the surrounding medium, and $A$ is the angle of the prism. When the prism is in air, $ \mu = \frac{3}{2} $ So, $ \delta = \left(\frac{3}{2} - 1\right)A $ $ \delta = \frac{1}{2}A $ Now the prism is immersed in water. The refractive index of glass with respect to water is $ \mu_{gw} = \frac{\mu_g}{\mu_w} $ $ \mu_{gw} = \frac{\frac{3}{2}}{\frac{4}{3}} $ $ \mu_{gw} = \frac{3}{2} \times \frac{3}{4} = \frac{9}{8} $ So the new angle of minimum deviation is $ \delta' = \left(\frac{9}{8} - 1\right)A $ $ \delta' = \frac{1}{8}A $ But $ \delta = \frac{1}{2}A $ So, $ \delta' = \frac{1}{4}\delta $ Hence, the angle of minimum deviation in water is $ \boxed{\frac{\delta}{4}} $ Correct option: C.

Correct Answer: (b)

Explanation: $\mu_{\text {water }}=1.33 \text { and } \mu_{\text {air }}=1$ As air is optically rarer, $\mu_{\text {air }}

Correct Answer: (a)

Explanation: To determine the focal length and type of lens used, we have the following information: Object distance, $ u = -25 \, \text{cm} $ (negative because it's on the opposite side of the light's direction), Image distance, $ v = +75 \, \text{cm} $. To find the focal length $ f $, we use the lens formula: $ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $ Substitute the given values: $ \frac{1}{f} = \frac{1}{75} - \frac{1}{-25} $ Simplify the equation: $ \frac{1}{f} = \frac{1}{75} + \frac{1}{25} $ Convert $ \frac{1}{25} $ to a denominator of 75: $ \frac{1}{25} = \frac{3}{75} $ Now, add the fractions: $ \frac{1}{f} = \frac{1}{75} + \frac{3}{75} = \frac{4}{75} $ To find $ f $, take the reciprocal: $ f = \frac{75}{4} = 18.75 \, \text{cm} $ Since the focal length $ f $ is positive, the lens is a convex lens.